Contents / Calculus / Limits & Continuity

Chapter 1

Limits & Continuity

Build intuition for approaching values and continuity.

Introduction

Limits are the foundation of all of calculus. Every derivative and every integral is secretly a limit. Before you can do anything else, you need to know what it means for a function to approach a value.

The notation $\lim_{x\to a} f(x) = L$ means: as $x$ gets arbitrarily close to $a$ , the outputs $f(x)$ get arbitrarily close to $L$ . Notice that we never require $f(a)$ itself to equal $L$ , or even to exist. That single omission is what makes limits powerful — they describe a function at points where the function has nothing to say.

Continuity is the special case where the limit and the actual value agree. Most functions you meet are continuous; the interesting calculus happens at the exceptions.

The chapter runs in the order a first course does: the families of functions calculus is built from, the two problems that force limits into existence, the informal and then the precise definition, the laws that let us compute limits without going back to the definition, continuity and the Intermediate Value Theorem, behaviour at infinity, and finally L'Hôpital's Rule, which uses the derivative to settle the limits that resist algebra.

1.1Functional representations: the language before the calculus

Calculus does not invent new functions. It studies how a handful of familiar families change, so fluency with those families comes first.

Linear functions $f(x) = mx + b$ model constant rates of change. Their graphs are straight lines, and that simplicity is exactly what a derivative reproduces locally: the tangent line is a linear approximation.

Power functions $f(x) = x^n$ produce polynomial behaviour. Quadratics model projectile motion, cubics bend and change concavity. Polynomials are the best-behaved functions in the subject: continuous, differentiable, no surprises. The degree of a polynomial, the highest power that appears, decides its behaviour far from the origin, and that fact returns in the section on limits at infinity.

Rational functions are quotients of polynomials, $f(x) = p(x)/q(x)$ . They inherit the good behaviour of polynomials everywhere except where $q(x) = 0$ , and those exceptional points are where the two most important kinds of bad behaviour in this chapter — holes and vertical asymptotes — live.

Exponential functions $f(x) = a^x$ — above all $e^x$ — model growth and decay: populations, radioactivity, compound interest. Their defining property is that the rate of growth is proportional to the current value, which is why $e^x$ turns out to be its own derivative.

Logarithmic functions $f(x) = \log_a(x)$ invert the exponentials. The natural logarithm $\ln x$ grows without bound, but astonishingly slowly, and it takes centre stage in integration through

\int \frac{1}{x}\,dx = \ln|x| + C.

Trigonometric functions $\sin x$ , $\cos x$ and $\tan x$ model anything periodic: waves, oscillations, circular motion. Their derivatives cycle through one another, and one of their limits,

\lim_{x\to 0} \frac{\sin x}{x} = 1,

underpins all of trigonometric calculus. It is proved later in this chapter.

Inverse functions swap input and output. If $f(x) = e^x$ then $f^{-1}(x) = \ln x$ ; for an inverse to exist at all, $f$ must be one-to-one. Derivatives of inverses obey

(f^{-1})'(x) = \frac{1}{f'\!\left(f^{-1}(x)\right)}.

Piecewise functions are defined by different rules on different intervals. They are the natural home of one-sided limits, of continuity and of differentiability — precisely the ideas that follow.

Intuition. Function families are like vehicle types. Linear functions are bicycles: steady, predictable speed. Exponentials are rockets: the faster they go, the faster they accelerate. Logarithms are the opposite — a rapid start, then diminishing returns. Trigonometric functions are Ferris wheels, going up and down forever in the same cycle.

Example 1.1 (Reading the domain off the formula). Find the domain of $f(x) = \dfrac{\sqrt{x-1}}{x-3}$ .

Solution. Two things can go wrong in a formula built from these families: a square root of a negative number and a division by zero. The root needs $x - 1 \ge 0$ , so $x \ge 1$ . The denominator needs $x \ne 3$ .

Both must hold at once, so the domain is $[1, 3) \cup (3, \infty)$ . The point $x = 3$ is exactly the kind of place a limit is designed to interrogate: the function says nothing there, but its neighbours do.□

Example 1.2 (Recognising a family from its rate of change). A colony of bacteria doubles every $3$ hours and starts at $5$ million cells. Which family models it, and how many cells are there after $9$ hours?

Solution. Doubling on a fixed schedule means the growth in each period is proportional to what is already present — the signature of the exponential family. Write

N(t) = 5 \cdot 2^{t/3}

with $t$ in hours: at $t = 3$ the exponent is $1$ and the population has doubled, as required. After $9$ hours the exponent is $3$ , so $N(9) = 5 \cdot 8 = 40$ million cells. Sanity check: three doublings of $5$ give $10$ , $20$ , $40$ .□

The derivative measures the speed of a function and the integral measures the total distance travelled. Both begin by recognising which family you are holding.

Exercises for §1.1 →

1.2The tangent and velocity problems

Two questions, one from geometry and one from physics, make limits unavoidable. Neither can be answered by ordinary algebra, and both have the same shape.

The tangent problem asks for the slope of the curve $y = x^{2}$ at the point $P = (1, 1)$ . A slope needs two points, and we have only one. The way out is to take a second point $Q = (x, x^{2})$ on the curve and compute the slope of the secant line $PQ$ :

m_{PQ} = \frac{x^2 - 1}{x - 1}.

For $x = 2$ this is $3$ ; for $x = 1.5$ it is $2.5$ ; for $x = 1.1$ it is $2.1$ ; for $x = 1.01$ it is $2.01$ . From the other side, $x = 0.9$ gives $1.9$ and $x = 0.99$ gives $1.99$ . The secant slopes close in on $2$ as $Q$ slides toward $P$ , and $2$ is the only reasonable candidate for the slope of the tangent.

Figure 1.3. Secants through

P=(1,1)

and

Q

y=x^{2}

. As

Q

slides from

(2,4)

(1.5,2.25)

the secant slopes fall from

3

2.5

, and the lines swing down towards the tangent of slope

2

. The tangent is what the secants approach, never one of them.

Definition 1.4 (Tangent line as a limit of secants). The tangent line to the curve $y = f(x)$ at the point $P = (a, f(a))$ is the line through $P$ with slope

m = \lim_{x\to a} \frac{f(x) - f(a)}{x - a},

provided this limit exists.

The formula cannot be evaluated by substituting $x = a$ : it reads $0/0$ . Everything in the definition hangs on the word limit, which is why a limit must be defined properly before the derivative can be.

The velocity problem is the same thing in different clothes. Drop a ball from a tower; Galileo's law says it has fallen $s(t) = 4.9t^{2}$ metres after $t$ seconds. What is its velocity at $t = 5$ exactly? Velocity is distance over time, but at a single instant no distance is covered and no time elapses. So measure the average velocity over a short interval $[5, 5+h]$ :

v_{\text{avg}} = \frac{s(5+h) - s(5)}{h} = \frac{4.9(5+h)^2 - 4.9 \cdot 25}{h}.

With $h = 1$ this is $53.9$ m/s; with $h = 0.1$ it is $49.49$ ; with $h = 0.01$ it is $49.049$ ; with $h = 0.001$ it is $49.0049$ . The averages settle on $49$ , and $49$ m/s is what we mean by the instantaneous velocity.

Definition 1.5 (Instantaneous velocity). If an object has position $s(t)$ at time $t$ , its instantaneous velocity at $t = a$ is

v(a) = \lim_{h\to 0} \frac{s(a+h) - s(a)}{h}.

Compare the two definitions: the velocity is the slope of the tangent to the position graph. That coincidence is the beginning of the derivative, and the next chapter takes it up. For now the lesson is that both problems reduce to one question — what does a quotient approach when both top and bottom head to zero? — and that question is the subject of this chapter.

Intuition. A speedometer does not measure speed at an instant, because it cannot: it measures how far the wheels turned over a very short time and divides. The reading is an average over that tiny interval. The instantaneous speed is what the readings converge to as the interval shrinks — a limit, computed by a piece of hardware.

Example 1.6 (Secant slopes closing on a tangent). Estimate the slope of the tangent to $y = x^{3}$ at $(2, 8)$ from secant slopes, then find it exactly.

Solution. The secant through $(2,8)$ and $(x, x^{3})$ has slope

m_{PQ} = \frac{x^3 - 8}{x - 2}.

At $x = 2.1$ this is $12.61$ ; at $x = 2.01$ it is $12.0601$ ; at $x = 1.99$ it is $11.9401$ . The slopes close in on $12$ .

To confirm, factor the numerator as a difference of cubes: $x^{3} - 8 = (x-2)(x^{2} + 2x + 4)$ . For $x \ne 2$ the factor $(x-2)$ cancels and $m_{PQ} = x^{2} + 2x + 4$ , which tends to $4 + 4 + 4 = 12$ as $x \to 2$ . The tangent line is $y - 8 = 12(x - 2)$ .□

Example 1.7 (Instantaneous velocity from average velocities). A stone is thrown upward so that its height after $t$ seconds is $s(t) = 20t - 5t^{2}$ metres. Find its velocity at $t = 1$ .

Solution. Average velocity over $[1, 1+h]$ is

\frac{s(1+h) - s(1)}{h} = \frac{20(1+h) - 5(1+h)^2 - 15}{h} = \frac{20 + 20h - 5 - 10h - 5h^2 - 15}{h} = \frac{10h - 5h^2}{h}.

For $h \ne 0$ this equals $10 - 5h$ , and as $h \to 0$ it tends to $10$ . The velocity at $t = 1$ is $10$ m/s upward. Sanity check: the stone starts at $20$ m/s and is slowing, so a value below $20$ and above $0$ is right; it peaks at $t = 2$ , where the same computation gives $0$ .□

Exercises for §1.2 →

1.3Building intuition: what does 'approaching' mean?

Take $f(x) = \frac{x^{2}-1}{x-1}$ and feed it values of $x$ marching toward $1$ : $0.9$ , $0.99$ , $0.999$ , then $1.001$ , $1.01$ , $1.1$ . The outputs are $1.9$ , $1.99$ , $1.999$ , $2.001$ , $2.01$ , $2.1$ . They cluster around $2$ , even though $f(1)$ is undefined. That clustering is the limit:

\lim_{x\to 1} \frac{x^2-1}{x-1} = 2.

Definition 1.8 (Limit of a function (informal)). We write $\lim_{x\to a} f(x) = L$ , and say the limit of $f(x)$ as $x$ approaches $a$ is $L$ , if the values of $f(x)$ can be made as close to $L$ as we like by taking $x$ sufficiently close to $a$ but not equal to $a$ .

Two phrases in the definition carry the weight. As close as we like means every tolerance, not just a small one: if someone demands $f(x)$ within one millionth of $L$ , we must be able to oblige. But not equal to $a$ means the value $f(a)$ never enters: it may be undefined, or defined and wrong, and the limit is unaffected. The limit cares about the neighbourhood around a point, never the point itself. That is exactly why it is useful: it lets us analyse behaviour where the function breaks.

The definition is informal because "as close as we like" and "sufficiently close" have not been pinned to numbers. The precise version, with $\varepsilon$ and $\delta$ , comes in its own section; the informal one is enough to compute with, and it is the one to hold in your head.

Intuition. Imagine walking toward a door, halving the remaining distance with every step. You never step through it — but you can get close enough to peek and see what is on the other side. That peek is the limit.

The door might be locked, or missing, or painted a surprising colour. None of that changes what you see through it from the corridor.

Example 1.9 (Building a limit table). Estimate $\lim_{x\to 2} \frac{x^{2} - 4}{x - 2}$ numerically, then confirm the value algebraically.

Solution. The function is undefined at $x = 2$ , where it reads $0/0$ , so approach from both sides.

From the left, $f(1.9) = 3.9$ , $f(1.99) = 3.99$ , $f(1.999) = 3.999$ . From the right, $f(2.1) = 4.1$ , $f(2.01) = 4.01$ , $f(2.001) = 4.001$ . The outputs approach $4$ from both sides.

Algebra confirms it: for $x \neq 2$ ,

\frac{x^2-4}{x-2} = \frac{(x-2)(x+2)}{x-2} = x+2,

and $\lim_{x\to 2}(x+2) = 4$ .□

Example 1.10 (A table that needs care). Estimate $\lim_{t\to 0} \dfrac{\sqrt{t^{2}+9} - 3}{t^{2}}$ .

Solution. A calculator gives $0.162278$ at $t = \pm 1$ , $0.165525$ at $t = \pm 0.5$ , $0.166620$ at $t = \pm 0.1$ and $0.166655$ at $t = \pm 0.05$ , which strongly suggests $1/6$ .

Push further, though, and the readings go wrong: a ten-digit calculator reports $0.168$ at $t = 0.0005$ , then $0.2$ at $t = 0.0001$ , then $0$ at $t = 0.00005$ . That is not the function changing its mind; it is rounding. When $t$ is tiny, $\sqrt{t^{2} + 9}$ is so close to $3$ that the calculator stores it as exactly $3$ , and the numerator collapses to $0$ .

The true value is $1/6$ . Multiplying top and bottom by $\sqrt{t^{2}+9}+3$ gives

\frac{(t^2 + 9) - 9}{t^2\left(\sqrt{t^2+9}+3\right)} = \frac{1}{\sqrt{t^2+9}+3} \longrightarrow \frac{1}{6}.

Numerical evidence suggests; algebra decides.□

Example 1.11 (When the values never settle). Investigate $\lim_{x\to 0} \sin\!\left(\dfrac{\pi}{x}\right)$ .

Solution. Try the obvious inputs: at $x = 1, \tfrac12, \tfrac13, \tfrac14, \ldots$ the argument $\pi/x$ is a multiple of $\pi$ , so every value is $0$ . A table built from these alone would announce a limit of $0$ .

It would be wrong. At $x = \tfrac{2}{5}, \tfrac{2}{9}, \tfrac{2}{13}, \ldots$ the argument is $\tfrac{5\pi}{2}, \tfrac{9\pi}{2}, \ldots$ and every value is $1$ ; at $x = \tfrac{2}{3}, \tfrac{2}{7}, \ldots$ every value is $-1$ . Arbitrarily close to $0$ the function takes every value between $-1$ and $1$ infinitely often. No single number $L$ can have all nearby outputs within, say, $\tfrac12$ of it, so the limit does not exist.

The graph oscillates infinitely fast near the origin. This example is the standard warning that a table samples the function and a sample can mislead.□

Pitfall. A table of values is evidence, not proof. Two things go wrong routinely: the sample points happen to miss the interesting behaviour, as with $\sin(\pi/x)$ ; and rounding error takes over when inputs get very small, as with $\sqrt{t^{2}+9}$ . Use tables to form a guess and algebra to confirm it.

Exercises for §1.3 →

1.4One-sided limits

Definition 1.12 (One-sided limits). The left-hand limit $\lim_{x\to a^-} f(x) = L$ means the values of $f(x)$ can be made as close to $L$ as we like by taking $x$ sufficiently close to $a$ with $x < a$ . The right-hand limit $\lim_{x\to a^+} f(x) = L$ is the same with $x > a$ .

The notation $x \to a^-$ is read " $x$ approaches $a$ from the left", and it says nothing about the sign of $a$ : $x \to (-3)^-$ means $x$ approaches $-3$ through values such as $-3.1$ and $-3.01$ .

Theorem 1.13 (Two-sided limits from one-sided limits). $\lim_{x\to a} f(x) = L$ if and only if both one-sided limits exist and $\lim_{x\to a^-} f(x) = \lim_{x\to a^+} f(x) = L$ .

Proof. Both directions come straight from the definition. A $\delta$ that works for the two-sided limit works for each side separately. Conversely, if $\delta_-$ and $\delta_+$ handle the two sides for a given $\varepsilon$ , then $\delta = \min(\delta_-, \delta_+)$ handles every $x$ with $0 < |x-a| < \delta$ .∎

The theorem is the standard tool for showing a limit fails to exist. For $f(x) = |x|/x$ we have $\lim_{x\to 0^-} f(x) = -1$ and $\lim_{x\to 0^+} f(x) = 1$ ; the one-sided limits disagree, so the two-sided limit at $0$ does not exist.

One-sided limits are also the right language at a domain boundary. For instance $\lim_{x\to 0^+} \ln x = -\infty$ , while the left-hand limit is meaningless because $\ln x$ is undefined for $x < 0$ . Likewise $\lim_{x\to 0^+}\sqrt{x} = 0$ is the only limit $\sqrt{x}$ has at the origin.

Intuition. Think of a one-way street. The left-hand limit is what you see driving in from the left, the right-hand limit what you see from the right. If both drivers report the same thing, the limit exists. If one sees a café and the other a parking lot, there is no single answer.

Example 1.14 (One-sided limits of a piecewise function). Evaluate the one-sided limits at $x = 3$ of

f(x) = \begin{cases} x+1, & x < 3 \\ 5, & x = 3 \\ x^2 - 5, & x > 3. \end{cases}

Solution.

Left-hand limit: the relevant piece is $x+1$ , so $\lim_{x\to 3^-}(x+1) = 4$ .
Right-hand limit: the relevant piece is $x^{2}-5$ , so $\lim_{x\to 3^+}(x^{2}-5) = 9 - 5 = 4$ .
The one-sided limits agree, so $\lim_{x\to 3} f(x) = 4$ by the two-sided criterion.
But $f(3) = 5 \neq 4$ , so $f$ has a removable discontinuity at $x = 3$ : the limit exists and simply fails to match the value.

□

Example 1.15 (An absolute value in disguise). Find $\lim_{x\to 2^-} \dfrac{x^{2} - 4}{|x - 2|}$ and $\lim_{x\to 2^+} \dfrac{x^{2} - 4}{|x - 2|}$ , and decide whether the two-sided limit exists.

Solution. The absolute value is the only obstacle, and it resolves differently on each side. For $x < 2$ we have $|x-2| = -(x-2) = 2 - x$ , so

\frac{x^2-4}{|x-2|} = \frac{(x-2)(x+2)}{-(x-2)} = -(x+2) \longrightarrow -4 \quad\text{as } x \to 2^-.

For $x > 2$ we have $|x-2| = x-2$ , so the quotient is $x + 2 \to 4$ as $x \to 2^+$ .

The one-sided limits are $-4$ and $4$ ; they differ, so $\lim_{x\to 2}$ does not exist. The graph has a jump of size $8$ at $x = 2$ .□

Example 1.16 (Reading one-sided limits from the floor function). Let $\lfloor x \rfloor$ be the greatest integer not exceeding $x$ . Find $\lim_{x\to 3^-}\lfloor x\rfloor$ , $\lim_{x\to 3^+}\lfloor x\rfloor$ and $\lim_{x\to 3.5}\lfloor x\rfloor$ .

Solution. Just to the left of $3$ , inputs such as $2.9$ and $2.999$ have floor $2$ , so the left-hand limit is $2$ . Just to the right, $3.001$ has floor $3$ , so the right-hand limit is $3$ . They disagree: no limit at $3$ , and the same happens at every integer.

At $3.5$ the function is constantly $3$ on the whole interval $(3, 4)$ , so both one-sided limits, and the limit, equal $3$ . The floor function is badly behaved only at the integers.□

Pitfall. A piecewise definition tells you which formula to use on each side, and the value assigned at the breakpoint is irrelevant to both one-sided limits. Students lose marks by substituting into the " $x = a$ " branch. That branch decides continuity, never the limit.

Exercises for §1.4 →

1.5Infinite limits and vertical asymptotes

Writing $\lim_{x\to a} f(x) = \infty$ says the outputs grow without bound as $x$ approaches $a$ . Strictly the limit does not exist as a real number; the symbol records how it fails. The infinity sign here is a description, not a number, and it cannot be substituted into the limit laws of the next sections.

Definition 1.17 (Infinite limit). $\lim_{x\to a} f(x) = \infty$ means the values of $f(x)$ can be made as large as we like by taking $x$ sufficiently close to $a$ but not equal to $a$ . Precisely: for every number $M$ there is a $\delta > 0$ such that

0 < |x-a| < \delta \quad\Longrightarrow\quad f(x) > M.

$\lim_{x\to a} f(x) = -\infty$ is defined the same way with $f(x) < M$ . One-sided versions restrict $x$ to one side of $a$ .

The $M$ plays the role a tolerance plays for finite limits: whatever height $M$ is demanded, a $\delta$ must push the graph above it. With $f(x) = 1/x^{2}$ and target $M = 10^6$ , any $x$ with $0 < |x| < 10^{-3}$ works; in general $\delta = 1/\sqrt{M}$ does the job, since $|x| < 1/\sqrt{M}$ gives $x^{2} < 1/M$ and $1/x^{2} > M$ .

Definition 1.18 (Vertical asymptote). The line $x = a$ is a vertical asymptote of $f$ if at least one of the one-sided limits of $f$ at $a$ is $+\infty$ or $-\infty$ .

For a rational function, asymptotes live where the denominator vanishes and the numerator does not, so factor the denominator first and cancel common factors. A cancelled factor is a hole, not a wall — this is the distinction the third example turns on.

Signs need care on each side separately: $\lim_{x\to 0^+} 1/x = +\infty$ while $\lim_{x\to 0^-} 1/x = -\infty$ . The reliable method is to write down, for $x$ just to one side of $a$ , the sign of every factor and then the sign of the quotient. When the two sides carry opposite signs, the graph escapes in both directions.

For $\ln x$ , $\tan x$ and $\csc x$ the asymptotes come from the nature of the function rather than from a visible denominator; $\tan x$ has one at every $x = \pm\pi/2, \pm 3\pi/2, \ldots$ and $\ln x$ has one at $x = 0$ from the right only.

Intuition. A vertical asymptote is an invisible wall the graph races toward but never reaches. To find the walls of a fraction, look where the denominator is zero. If the numerator is nonzero there, you have a wall; if it is also zero, the factors may cancel and leave a pothole instead.

Example 1.19 (Signs on the two sides of an asymptote). Find $\lim_{x\to 3^+}\dfrac{2x}{x-3}$ and $\lim_{x\to 3^-}\dfrac{2x}{x-3}$ .

Solution. Substituting $x = 3$ gives $6/0$ : nonzero over zero, so the function blows up and only the sign remains to be found.

For $x$ slightly larger than $3$ , say $3.01$ , the numerator is near $6 > 0$ and the denominator is $0.01 > 0$ , so the quotient is large and positive: $\lim_{x\to 3^+} = +\infty$ .

For $x$ slightly smaller, say $2.99$ , the numerator is still near $6$ but the denominator is $-0.01 < 0$ , so the quotient is large and negative: $\lim_{x\to 3^-} = -\infty$ . The line $x = 3$ is a vertical asymptote, approached upward from the right and downward from the left.□

Example 1.20 (Proving an infinite limit from the definition). Show from the definition that $\lim_{x\to 0} \dfrac{1}{x^{2}} = \infty$ , and find a $\delta$ that works for $M = 400$ .

Solution. Given $M > 0$ , we need $\delta$ with $1/x^{2} > M$ whenever $0 < |x| < \delta$ . Since $1/x^{2} > M$ is equivalent to $x^{2} < 1/M$ , that is, $|x| < 1/\sqrt{M}$ , the choice $\delta = 1/\sqrt{M}$ works: if $0 < |x| < 1/\sqrt{M}$ then $x^{2} < 1/M$ and $1/x^{2} > M$ .

For $M = 400$ this gives $\delta = 1/20 = 0.05$ . Check: $x = 0.04$ gives $1/x^{2} = 625 > 400$ .□

Example 1.21 (Vertical asymptote or hole?). Find the vertical asymptotes of $f(x) = \frac{x^{2}-1}{x^{2}-x}$ .

Solution.

Factor: $f(x) = \frac{(x-1)(x+1)}{x(x-1)}$ , so the denominator vanishes at $x=0$ and $x=1$ .
At $x=1$ the factor $(x-1)$ cancels, leaving $\frac{x+1}{x}$ for $x \neq 1$ . That is a hole, and the limit there is $\frac{2}{1} = 2$ .
At $x=0$ the reduced numerator is $1 \neq 0$ while the denominator vanishes, so $x = 0$ is a vertical asymptote.
The two sides differ in sign: $\lim_{x\to 0^+}\frac{x+1}{x} = +\infty$ and $\lim_{x\to 0^-}\frac{x+1}{x} = -\infty$ .

□

Example 1.22 (An asymptote without a denominator). Find $\lim_{x\to (\pi/2)^-} \tan x$ and $\lim_{x\to (\pi/2)^+} \tan x$ .

Solution. Write $\tan x = \sin x/\cos x$ . As $x \to \pi/2$ the numerator tends to $1$ while $\cos x \to 0$ , so the quotient blows up. Just left of $\pi/2$ the cosine is positive and small, giving $+\infty$ ; just right of it the cosine is negative and small, giving $-\infty$ . The line $x = \pi/2$ is a vertical asymptote of $\tan x$ , and the same argument places one at every odd multiple of $\pi/2$ .□

Pitfall. "Denominator zero" is not the test for a vertical asymptote; "denominator zero and numerator nonzero after cancelling" is. Both $\frac{x^{2}-1}{x-1}$ and $\frac{1}{x-1}$ have a zero denominator at $x = 1$ ; only the second has an asymptote there.

Exercises for §1.5 →

1.6The precise definition of a limit

The informal definition uses the words "as close as we like" and "sufficiently close". To prove anything about limits — that the limit laws hold, that a limit is unique, that a particular limit is what a table suggests — those words have to become inequalities.

Definition 1.23 (Limit of a function). Let $f$ be defined on an open interval containing $a$ , except possibly at $a$ itself. We write $\lim_{x\to a} f(x) = L$ when for every $\varepsilon > 0$ there exists a $\delta > 0$ such that

0 < |x - a| < \delta \quad\Longrightarrow\quad |f(x) - L| < \varepsilon.

Read the definition as a game between a challenger and you. The challenger names an output tolerance $\varepsilon$ , as small as they please. You must answer with an input tolerance $\delta$ so that every $x$ within $\delta$ of $a$ (other than $a$ itself) has its output within $\varepsilon$ of $L$ . The limit is $L$ if you can win for every $\varepsilon$ ; it fails to be $L$ if the challenger can pick one $\varepsilon$ for which no $\delta$ works.

Notice that $\delta$ is allowed to depend on $\varepsilon$ — a tighter output demand will usually need a tighter input — and that once a $\delta$ works, every smaller $\delta$ works too. The clause $0 < |x-a|$ is what excludes the point $a$ itself.

A proof from the definition always has the same two halves. First, a private calculation: start from $|f(x) - L| < \varepsilon$ and work backwards to see how small $|x - a|$ must be. Second, the proof proper: announce the $\delta$ found, and verify forwards that it delivers $|f(x) - L| < \varepsilon$ .

Intuition. A machinist is told a shaft must be $50$ mm in diameter to within $\varepsilon = 0.01$ mm. She knows the lathe setting $x$ produces diameter $f(x)$ . The question she actually has to answer is: how precisely must I set the dial, so that any setting within $\delta$ of the right one yields a shaft within tolerance? If a tighter tolerance always has an answer, the diameter is a continuous function of the dial — and the limit is what the dial promises.

Example 1.24 (An ε–δ proof for a linear function). Prove that $\lim_{x\to 3}(4x - 5) = 7$ .

Solution. Preliminary analysis. We want $|(4x - 5) - 7| < \varepsilon$ , that is $|4x - 12| < \varepsilon$ , that is $4|x - 3| < \varepsilon$ , that is $|x - 3| < \varepsilon/4$ . So $\delta = \varepsilon/4$ should work.

Proof. Let $\varepsilon > 0$ be given and set $\delta = \varepsilon/4$ . If $0 < |x - 3| < \delta$ , then

|(4x-5) - 7| = |4x - 12| = 4|x-3| < 4 \cdot \frac{\varepsilon}{4} = \varepsilon.

Hence $\lim_{x\to 3}(4x-5) = 7$ . For a line of slope $m$ the pattern is always $\delta = \varepsilon/|m|$ : a steep line converts a small input error into a larger output error, so it needs a smaller $\delta$ .□

Example 1.25 (An ε–δ proof for a quadratic). Prove that $\lim_{x\to 3} x^{2} = 9$ .

Solution. Preliminary analysis. We want $|x^{2} - 9| = |x - 3|\,|x + 3| < \varepsilon$ . The factor $|x-3|$ we control directly, but $|x+3|$ varies with $x$ . The remedy is to agree in advance to keep $x$ within $1$ of $3$ : if $|x - 3| < 1$ then $2 < x < 4$ , so $5 < x + 3 < 7$ and $|x+3| < 7$ . Under that agreement $|x^{2} - 9| < 7|x - 3|$ , which is below $\varepsilon$ as soon as $|x - 3| < \varepsilon/7$ .

Proof. Let $\varepsilon > 0$ and set $\delta = \min(1, \varepsilon/7)$ . If $0 < |x-3| < \delta$ , then $|x - 3| < 1$ so $|x+3| < 7$ , and also $|x-3| < \varepsilon/7$ , so

|x^2 - 9| = |x-3|\,|x+3| < \frac{\varepsilon}{7}\cdot 7 = \varepsilon.

The $\min$ is the standard device for non-linear functions: one bound tames the awkward factor, the other delivers the tolerance. For $\varepsilon = 0.5$ the recipe gives $\delta = 1/14$ .□

Example 1.26 (Using the definition to refute a limit). Show that $\lim_{x\to 0} \dfrac{|x|}{x}$ does not exist.

Solution. Suppose it existed and equalled some $L$ . Take $\varepsilon = 1$ . Whatever $\delta > 0$ is offered, the point $x = \delta/2$ has $f(x) = 1$ and the point $x = -\delta/2$ has $f(x) = -1$ , both within $\delta$ of $0$ . So $|1 - L| < 1$ and $|-1 - L| < 1$ , which force $L > 0$ and $L < 0$ at once. No $L$ survives, so the limit does not exist. In the language of the game, the challenger wins with $\varepsilon = 1$ against every $\delta$ .□

Proposition 1.27 (Uniqueness of limits). If $\lim_{x\to a} f(x) = L$ and $\lim_{x\to a} f(x) = M$ , then $L = M$ .

Proof. Suppose $L \ne M$ and take $\varepsilon = |L - M|/2$ . Choose $\delta$ small enough that both $|f(x) - L| < \varepsilon$ and $|f(x) - M| < \varepsilon$ whenever $0 < |x-a| < \delta$ ; then for such an $x$ the triangle inequality gives $|L - M| \le |L - f(x)| + |f(x) - M| < 2\varepsilon = |L - M|$ , a contradiction.∎

Pitfall. In a proof, $\delta$ must be chosen before $x$ and may depend only on $\varepsilon$ (and on $a$ and $f$ ). A "proof" that picks $\delta$ after looking at a particular $x$ proves nothing. Likewise the analysis half, working backwards from $\varepsilon$ , is scratch work; the argument the reader must accept is the forward verification.

Exercises for §1.6 →

1.7The limit laws

Going back to the $\varepsilon$ – $\delta$ definition for every limit would be unbearable. The limit laws let us build complicated limits from simple ones, and they are proved once, from the definition, so that nobody has to do it again.

Theorem 1.28 (Limit Laws). Suppose $c$ is a constant and the limits $\lim_{x\to a} f(x) = L$ and $\lim_{x\to a} g(x) = M$ both exist. Then

\lim_{x\to a}\,[f(x) + g(x)] = L + M, \qquad \lim_{x\to a}\,[f(x) - g(x)] = L - M, \qquad \lim_{x\to a}\,[c\,f(x)] = cL,

\lim_{x\to a}\,[f(x)\,g(x)] = LM, \qquad \lim_{x\to a}\frac{f(x)}{g(x)} = \frac{L}{M} \ \text{ provided } M \ne 0,

\lim_{x\to a}\,[f(x)]^n = L^n, \qquad \lim_{x\to a}\sqrt[n]{f(x)} = \sqrt[n]{L}

for every positive integer $n$ , with $L > 0$ required in the root law when $n$ is even.

In words: the limit of a sum is the sum of the limits, and likewise for differences, constant multiples, products, quotients, powers and roots. Every law carries the same hypothesis — the individual limits must exist — and the quotient law adds that the limit of the denominator is nonzero.

Proof. Sum law. Let $\varepsilon > 0$ . Since $f \to L$ , there is $\delta_1$ with $|f(x) - L| < \varepsilon/2$ whenever $0 < |x-a| < \delta_1$ ; since $g \to M$ , there is $\delta_2$ with $|g(x) - M| < \varepsilon/2$ whenever $0 < |x-a| < \delta_2$ . Put $\delta = \min(\delta_1, \delta_2)$ . If $0 < |x-a| < \delta$ then both estimates hold and, by the triangle inequality,

|[f(x) + g(x)] - (L + M)| \le |f(x) - L| + |g(x) - M| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.

The difference law follows from the sum law and the constant-multiple law with $c = -1$ ; the constant-multiple law is the sum law's argument with $|c|\,|f(x) - L| < \varepsilon$ obtained from $|f(x) - L| < \varepsilon/(|c| + 1)$ .

Product law, the idea. Write

f(x)g(x) - LM = f(x)\,[g(x) - M] + M\,[f(x) - L].

The second term is small because $f \to L$ . The first is small because $g \to M$ and, once $x$ is close to $a$ , $|f(x)| \le |L| + 1$ stays bounded; a bounded quantity times a quantity tending to zero tends to zero. The quotient law reduces to the product law once one shows $1/g(x) \to 1/M$ , which uses $M \ne 0$ to keep $|g(x)| > |M|/2$ near $a$ . The power law is the product law applied $n$ times; the root law needs a separate argument that belongs to real analysis.∎

Two consequences are used constantly. Since $\lim_{x\to a} c = c$ and $\lim_{x\to a} x = a$ are immediate from the definition (take $\delta = \varepsilon$ for the second), repeated use of the sum, constant-multiple and power laws gives the following.

Corollary 1.29 (Direct substitution property). If $p$ is a polynomial then $\lim_{x\to a} p(x) = p(a)$ . If $f = p/q$ is a rational function and $q(a) \ne 0$ then $\lim_{x\to a} f(x) = f(a)$ .

Proposition 1.30 (Limits preserve inequalities). If $f(x) \le g(x)$ for all $x$ near $a$ (except possibly at $a$ ) and both limits exist, then $\lim_{x\to a} f(x) \le \lim_{x\to a} g(x)$ .

Proof. Suppose instead $L > M$ . Taking $\varepsilon = (L - M)/2$ and $x$ close enough to $a$ makes $f(x) > L - \varepsilon = M + \varepsilon > g(x)$ , contradicting $f(x) \le g(x)$ .∎

Note what the proposition does not say: a strict inequality $f(x) < g(x)$ does not give a strict inequality of limits. With $f(x) = 0$ and $g(x) = x^{2}$ , $f < g$ for all $x \ne 0$ , yet both limits at $0$ are $0$ .

Intuition. The limit laws are the reason limits are computable at all. A statement like $\lim_{x\to 2}(x^{3} - 3x + 1) = 3$ hides eight applications of the laws — three products, a constant multiple, two sums and two evaluations of $\lim x = 2$ — and none of them is ever written out, because the corollary packages the whole thing as "substitute".

Example 1.31 (Combining limits from given values). Suppose $\lim_{x\to a} f(x) = 4$ and $\lim_{x\to a} g(x) = -2$ . Find $\lim_{x\to a}\dfrac{f(x)^2 - 3g(x)}{f(x) + g(x)}$ and $\lim_{x\to a}\sqrt{f(x) - 5g(x)}$ .

Solution. For the first, the denominator tends to $4 + (-2) = 2 \ne 0$ , so the quotient law applies. Numerator: by the power and constant-multiple laws, $4^2 - 3(-2) = 16 + 6 = 22$ . The limit is $22/2 = 11$ .

For the second, the expression under the root tends to $4 - 5(-2) = 14 > 0$ , so the root law applies and the limit is $\sqrt{14}$ .□

Example 1.32 (When the laws are silent). Let $f(x) = |x|/x$ and $g(x) = -|x|/x$ . Discuss $\lim_{x\to 0}[f(x) + g(x)]$ .

Solution. Neither $\lim_{x\to 0} f(x)$ nor $\lim_{x\to 0} g(x)$ exists, so the sum law cannot be invoked. But $f(x) + g(x) = 0$ for every $x \ne 0$ , and the limit of the sum is $0$ . The laws are one-directional: when the pieces have limits, the whole has a limit computed from them; when a piece fails, the whole may or may not have a limit, and the laws say nothing. You must look at the combined expression directly.□

Example 1.33 (Substitution is a theorem, not a reflex). Compute $\lim_{x\to -1}\dfrac{x^{3} - 2x + 5}{x^{2} + 3}$ .

Solution. This is a rational function and the denominator at $x = -1$ is $1 + 3 = 4 \ne 0$ , so the direct substitution property applies:

\frac{(-1)^3 - 2(-1) + 5}{(-1)^2 + 3} = \frac{-1 + 2 + 5}{4} = \frac{6}{4} = \frac{3}{2}.

Sanity check: at $x = -0.9$ the quotient is $6.071/3.81 \approx 1.59$ , close to $1.5$ and drifting toward it.□

Pitfall. The quotient law needs the limit of the denominator to be nonzero, and the product law needs both limits to exist. The forms $0/0$ and $0 \cdot \infty$ are precisely the cases where the laws refuse to answer; the sections that follow are about what to do then.

Exercises for §1.7 →

1.8Evaluating limits: direct substitution

Always try substitution first. If $f$ is continuous at $a$ then $\lim_{x\to a} f(x) = f(a)$ and you are done. The direct substitution property covers polynomials and rational functions; the continuity section extends the same conclusion to roots, exponentials, logarithms, trigonometric functions and their inverses, and to any sum, product, quotient or composition of these, on their domains.

Substitution fails in exactly two interesting ways, and they mean opposite things.

Indeterminate form: substitution returns $0/0$ or $\infty/\infty$ . The limit may still exist, but the expression as written carries too little information to decide; you must simplify or apply L'Hôpital's Rule.

Nonzero over zero: substitution returns something like $5/0$ . Then the function blows up and the limit is not a finite number — you are looking at a vertical asymptote, and the only remaining question is the sign on each side.

Intuition. Direct substitution asks: what happens if I just plug the number in? For well-behaved functions nothing strange happens at the point, so you simply evaluate.

It only breaks when the answer is meaningless, like $0/0$ . That is the function telling you: it is not that simple here, dig deeper.

Example 1.34 (When substitution works and when it fails). Decide what substitution tells you about $\lim_{x\to 2}(x^{3}+1)$ , $\lim_{x\to 0} \frac{\sin x}{x}$ and $\lim_{x\to 3} \frac{x+1}{x-3}$ .

Solution.

$\lim_{x\to 2}(x^{3}+1) = 2^3 + 1 = 9$ . Polynomials are continuous everywhere, so substitution is the whole argument.
$\frac{\sin 0}{0} = \frac{0}{0}$ is indeterminate; substitution decides nothing. Another technique is needed, and the limit turns out to equal $1$ .
$\frac{4}{0}$ is not indeterminate — the numerator is nonzero. The function blows up, and $x = 3$ is a vertical asymptote.

□

Example 1.35 (Substitution through a composition). Compute $\lim_{x\to 4}\sqrt{x^{2} + 9}$ and $\lim_{x\to \pi/4}\dfrac{\sin x}{1 + \cos x}$ .

Solution. The polynomial $x^{2} + 9$ tends to $25$ , and the root law (equivalently, continuity of $\sqrt{\ }$ at $25$ ) gives $\sqrt{25} = 5$ .

In the second, sine and cosine are continuous, the denominator tends to $1 + \frac{\sqrt2}{2} \ne 0$ , and the quotient law gives

\frac{\sqrt2/2}{1 + \sqrt2/2} = \frac{\sqrt2}{2 + \sqrt2} = \frac{\sqrt2(2 - \sqrt2)}{2} = \sqrt2 - 1.

Sanity check: $\sqrt2 - 1 \approx 0.414$ , and $0.707/1.707 \approx 0.414$ .□

Example 1.36 (Substitution after a sign check). Compute $\lim_{x\to 1}\dfrac{x^{2} + x - 2}{x^{2} - 3x + 2}$ .

Solution. Substituting $x = 1$ gives $0/0$ . Substitution has failed, but it has failed in the informative way: both polynomials vanish at $1$ , so both have $(x - 1)$ as a factor. Indeed $x^{2} + x - 2 = (x-1)(x+2)$ and $x^{2} - 3x + 2 = (x-1)(x-2)$ , so for $x \ne 1$ the quotient is $(x+2)/(x-2)$ , and substitution now gives $3/(-1) = -3$ .

The next section is about this move and its relatives.□

Pitfall. Substituting and obtaining a number is not always the end of the story: it is the end only when the function is continuous there. For $f(x) = \lfloor x \rfloor$ substitution gives $f(2) = 2$ , but $\lim_{x\to 2} f(x)$ does not exist. Substitution is licensed by continuity, and piecewise-defined functions are where the licence lapses.

Exercises for §1.8 →

1.9Indeterminate forms and algebraic tricks

The symbol $0/0$ is called indeterminate because the limit behind it could be anything. Compare three quotients, all of form $0/0$ at $x = 0$ :

\frac{x^2}{x} \to 0, \qquad \frac{5x}{x} \to 5, \qquad \frac{x}{x^3} \to \infty.

The form records that both parts vanish; it does not record how fast, and the rate is the whole question. The cure is always the same in spirit — rewrite the expression until substitution becomes legal — and the rewriting is licensed by one observation.

Theorem 1.37 (Limits ignore the point itself). If $f(x) = g(x)$ for all $x$ near $a$ except possibly at $a$ , and $\lim_{x\to a} g(x) = L$ , then $\lim_{x\to a} f(x) = L$ .

Proof. The $\varepsilon$ – $\delta$ condition for $g$ involves only values at points $x$ with $0 < |x - a| < \delta$ , and at every such point $f$ and $g$ agree (shrinking $\delta$ if necessary so that the agreement covers the whole punctured interval). So the same $\delta$ witnesses the condition for $f$ .∎

That is why cancelling $(x - a)$ is legitimate: the cancelled expression is a different function, but it agrees with the original everywhere except at $a$ , and the limit cannot tell them apart. Four rewrites cover almost every exercise.

Factor and cancel: when numerator and denominator share a factor $(x-a)$ , cancel it and substitute again.

Rationalise: when a square root appears, multiply by the conjugate. For $\frac{\sqrt{x+1}-1}{x}$ , multiply above and below by $\sqrt{x+1}+1$ .

Combine fractions: when two fractions are being subtracted, put them over a common denominator first. This is the standard treatment of the $\infty - \infty$ form at a finite point.

Use identities: $\sin^{2} x + \cos^{2} x = 1$ , the double-angle formulas, or $1-\cos x = 2\sin^{2}(x/2)$ often collapse the whole expression.

Intuition. An indeterminate form is the mathematical equivalent of "I don't know yet". Two runners both decelerate to a standstill at the finishing line; asking which was faster at the line is not answered by the fact that both stopped. You have to compare the rates at which they slowed, and algebra — later, differentiation — is how you extract those rates.

Example 1.38 (Factoring a removable discontinuity). Compute $\lim_{x\to 3} \frac{x^{2}-9}{x-3}$ .

Solution.

Substitution gives $\frac{9-9}{3-3} = \frac{0}{0}$ : indeterminate, so simplify.
The numerator is a difference of squares, $x^{2} - 9 = (x-3)(x+3)$ .
Since $x \to 3$ but $x \neq 3$ , the factor $(x-3)$ may be cancelled, leaving $\lim_{x\to 3}(x+3)$ .
Substitution is now legal: the limit is $6$ .

The graph has a hole at $x = 3$ ; the limit sees straight through it to the value the function approaches.□

Example 1.39 (Rationalising a radical expression). Compute $\lim_{x\to 0} \frac{\sqrt{1+x} - 1}{x}$ .

Solution.

Substitution gives $\frac{\sqrt{1}-1}{0} = \frac{0}{0}$ : indeterminate.
The obstacle is the root, so multiply by the conjugate $\frac{\sqrt{1+x}+1}{\sqrt{1+x}+1}$ .
The numerator becomes $(\sqrt{1+x})^2 - 1^2 = x$ , so the quotient is $\frac{x}{x(\sqrt{1+x}+1)} = \frac{1}{\sqrt{1+x}+1}$ .
Substituting $x=0$ gives $\frac{1}{2}$ .

The conjugate trick is the standard move whenever a limit contains $\sqrt{\text{something}} - \text{number}$ .□

Example 1.40 (Combining fractions: an ∞ − ∞ form). Compute $\lim_{x\to 1}\left(\dfrac{1}{x - 1} - \dfrac{2}{x^{2} - 1}\right)$ .

Solution.

Each piece blows up as $x \to 1$ , one to $+\infty$ and the other likewise, so the expression has the form $\infty - \infty$ , which is indeterminate: the two infinities may or may not cancel.
Put them over the common denominator $x^{2} - 1 = (x-1)(x+1)$ :

\frac{1}{x-1} - \frac{2}{(x-1)(x+1)} = \frac{(x+1) - 2}{(x-1)(x+1)} = \frac{x - 1}{(x-1)(x+1)}.

Cancel the common factor: for $x \ne 1$ the expression is $\frac{1}{x+1}$ .
Substitution now gives $\frac12$ . The infinities cancelled exactly, leaving a finite value.

□

Example 1.41 (A difference quotient). Compute $\lim_{h\to 0}\dfrac{(3+h)^{-1} - 3^{-1}}{h}$ .

Solution.

Substituting $h = 0$ gives $0/0$ .
Simplify the numerator first: $\frac{1}{3+h} - \frac13 = \frac{3 - (3+h)}{3(3+h)} = \frac{-h}{3(3+h)}$ .
Divide by $h$ , that is multiply by $1/h$ : for $h \ne 0$ the quotient is $\frac{-1}{3(3+h)}$ .
Letting $h \to 0$ gives $-\frac19$ .

This is the derivative of $1/x$ at $x = 3$ , and indeed $-1/x^{2}$ at $x = 3$ is $-1/9$ .□

Example 1.42 (A root in the numerator and a factorable denominator). Compute $\lim_{x\to 4}\dfrac{\sqrt{x} - 2}{x - 4}$ in two ways.

Solution. Conjugate method. Multiply above and below by $\sqrt x + 2$ :

\frac{(\sqrt x - 2)(\sqrt x + 2)}{(x-4)(\sqrt x + 2)} = \frac{x - 4}{(x-4)(\sqrt x + 2)} = \frac{1}{\sqrt x + 2} \longrightarrow \frac14.

Substitution method. Put $u = \sqrt x$ , so $x = u^{2}$ and $u \to 2$ as $x \to 4$ . Then

\frac{u - 2}{u^2 - 4} = \frac{u-2}{(u-2)(u+2)} = \frac{1}{u+2} \longrightarrow \frac14.

Two routes, one answer — a useful habit for checking work.□

Pitfall. Cancelling changes the function's domain, and the cancelled expression is only equal to the original away from the point. Writing $\frac{x^{2}-9}{x-3} = x + 3$ without the proviso $x \ne 3$ is false at $x = 3$ ; writing $\lim_{x\to 3}\frac{x^{2}-9}{x-3} = \lim_{x\to 3}(x+3)$ is correct, by the theorem above. Keep the limit sign attached while you cancel.

Exercises for §1.9 →

1.10The squeeze theorem

Theorem 1.43 (Squeeze Theorem). If $g(x) \le f(x) \le h(x)$ for all $x$ near $a$ (except possibly at $a$ ), and

\lim_{x\to a} g(x) = \lim_{x\to a} h(x) = L,

then $\lim_{x\to a} f(x) = L$ .

Proof. Let $\varepsilon > 0$ . Choose $\delta_1$ so that $|g(x) - L| < \varepsilon$ and $\delta_2$ so that $|h(x) - L| < \varepsilon$ on the respective punctured neighbourhoods, and let $\delta$ be the smaller of these (shrunk further if necessary so that the inequality $g \le f \le h$ holds throughout). For $0 < |x - a| < \delta$ we then have

L - \varepsilon < g(x) \le f(x) \le h(x) < L + \varepsilon,

so $|f(x) - L| < \varepsilon$ .∎

The theorem is the tool of choice when a factor oscillates without settling. The oscillation is bounded, a second factor shrinks to zero, and the product is dragged to zero with it — an argument no amount of substitution or factoring can supply, because the oscillating factor has no limit at all.

Note the hypothesis "for all $x$ near $a$ ": the inequalities need not hold everywhere, only on some punctured interval around $a$ , and what happens at $a$ itself is irrelevant as always.

Corollary 1.44 (Zero times bounded). If $\lim_{x\to a} f(x) = 0$ and $|g(x)| \le K$ for all $x$ near $a$ , then $\lim_{x\to a} f(x)g(x) = 0$ .

Proof. $-K|f(x)| \le f(x)g(x) \le K|f(x)|$ , and both bounds tend to $0$ since $|f(x)| \to 0$ . Apply the Squeeze Theorem.∎

Intuition. You are walking between two friends on a pavement, and both are heading for the same coffee shop. Stuck between them, you end up at the coffee shop too. You have no choice — and nobody needs to know anything about your own route.

Example 1.45 (Squeezing an oscillating function). Compute $\lim_{x\to 0} x^{2} \cos(1/x)$ .

Solution.

Direct substitution is hopeless: $\cos(1/x)$ oscillates between $-1$ and $1$ ever faster as $x\to 0$ , and $\cos(1/0)$ is meaningless.
Bound the oscillation instead: $-1 \le \cos(1/x) \le 1$ for every $x \neq 0$ .
Multiply through by $x^{2} > 0$ , which preserves the inequalities: $-x^{2} \le x^{2}\cos(1/x) \le x^{2}$ .
Both outer functions tend to $0$ as $x\to 0$ , so the Squeeze Theorem gives $\lim_{x\to 0} x^{2}\cos(1/x) = 0$ .

The chaos never had to be understood — only bounded. The factor $x^{2}$ tames it.□

Example 1.46 (A squeeze with an exponential). Show that $\lim_{x\to 0^+} \sqrt{x}\,e^{\sin(1/x)} = 0$ .

Solution.

The exponent $\sin(1/x)$ lies between $-1$ and $1$ , so $e^{\sin(1/x)}$ lies between $e^{-1}$ and $e$ — bounded, though never close to a single value.
For $x > 0$ multiply by $\sqrt x$ : $\;0 < \sqrt{x}\,e^{-1} \le \sqrt{x}\,e^{\sin(1/x)} \le \sqrt{x}\,e$ .
Both outer expressions tend to $0$ as $x \to 0^+$ .
By the Squeeze Theorem the limit is $0$ .

□

Example 1.47 (Squeezing a sequence-like expression). Find $\lim_{x\to\infty}\dfrac{\cos x}{x}$ .

Solution. The numerator has no limit at infinity, so the quotient law is unavailable. But $-1 \le \cos x \le 1$ gives, for $x > 0$ ,

-\frac{1}{x} \le \frac{\cos x}{x} \le \frac{1}{x},

and both bounds tend to $0$ as $x \to \infty$ . The Squeeze Theorem (which holds verbatim for limits at infinity) gives $0$ . The graph oscillates forever but inside an envelope that closes on the axis, so $y = 0$ is a horizontal asymptote crossed infinitely often.□

Pitfall. The bounds must actually sandwich the function on a whole punctured neighbourhood, and they must have the same limit. From $-1 \le \sin(1/x) \le 1$ alone nothing follows about $\lim_{x\to 0}\sin(1/x)$ : the bounds have limits $-1$ and $1$ , which differ, and in fact that limit does not exist. Also watch the sign when multiplying an inequality: multiplying by a negative quantity reverses it, which is why $x^{2}$ and $|x|$ , never $x$ , appear as the squeezing factor.

Exercises for §1.10 →

1.11Key trigonometric limits

These limits appear constantly and are worth committing to memory.

\lim_{x\to 0} \frac{\sin x}{x} = 1, \qquad \lim_{x\to 0} \frac{1-\cos x}{x} = 0, \qquad \lim_{x\to 0} \frac{\tan x}{x} = 1, \qquad \lim_{x\to 0} \frac{e^x - 1}{x} = 1.

The first is the load-bearing one, and it deserves a proof, because it cannot be obtained from the limit laws: the form is $0/0$ and no algebraic identity removes it. The proof is geometric, and it is where the Squeeze Theorem earns its keep.

Theorem 1.48 (The fundamental trigonometric limit). With $x$ measured in radians,

\lim_{x\to 0} \frac{\sin x}{x} = 1.

Proof. Take $0 < x < \pi/2$ and work in the unit circle. Let $O$ be the centre, $A = (1,0)$ , and let $B$ be the point at angle $x$ , so that $B = (\cos x, \sin x)$ . Let $T$ be the point where the tangent line at $A$ meets the ray $OB$ , so $T = (1, \tan x)$ .

Compare three areas. The triangle $OAB$ has base $OA = 1$ and height $\sin x$ , so its area is $\tfrac12\sin x$ . The circular sector $OAB$ has area $\tfrac12 x$ , since a sector of angle $x$ in a unit circle has area $x/2$ (this is where radians are essential). The triangle $OAT$ has base $1$ and height $\tan x$ , so its area is $\tfrac12\tan x$ . The triangle is contained in the sector, which is contained in the larger triangle, so

\frac12 \sin x \;\le\; \frac12 x \;\le\; \frac12 \tan x.

Multiply by $2$ and divide by $\sin x > 0$ :

1 \le \frac{x}{\sin x} \le \frac{1}{\cos x}.

Take reciprocals, which reverses the inequalities:

\cos x \le \frac{\sin x}{x} \le 1.

As $x \to 0^+$ we have $\cos x \to 1$ , so the Squeeze Theorem gives $\lim_{x\to 0^+}\frac{\sin x}{x} = 1$ . Finally $\frac{\sin x}{x}$ is an even function — replacing $x$ by $-x$ changes the sign of both numerator and denominator — so the left-hand limit equals the right-hand limit, and the two-sided limit is $1$ .∎

Corollary 1.49 (The cosine limit).

\lim_{x\to 0}\frac{\cos x - 1}{x} = 0, \qquad\text{equivalently}\qquad \lim_{x\to 0}\frac{1 - \cos x}{x} = 0.

Proof. Multiply by the conjugate $\cos x + 1$ :

\frac{\cos x - 1}{x} = \frac{\cos^2 x - 1}{x(\cos x + 1)} = \frac{-\sin^2 x}{x(\cos x+1)} = -\frac{\sin x}{x}\cdot\frac{\sin x}{\cos x + 1}.

As $x \to 0$ the first factor tends to $1$ and the second to $0/2 = 0$ , so the product tends to $0$ .∎

The tangent limit follows from $\frac{\tan x}{x} = \frac{\sin x}{x}\cdot\frac{1}{\cos x} \to 1\cdot 1$ . A companion worth knowing is

\lim_{x\to 0}\frac{1 - \cos x}{x^2} = \frac12,

which the same conjugate trick gives: the expression equals $\left(\frac{\sin x}{x}\right)^2\frac{1}{1 + \cos x} \to 1 \cdot \frac12$ .

Proposition 1.50 (Generalised sine limit). For constants $k$ and $m \neq 0$ , $\displaystyle\lim_{x\to 0} \frac{\sin(kx)}{mx} = \frac{k}{m}$ .

Proof. Write $\frac{\sin(kx)}{mx} = \frac{k}{m}\cdot\frac{\sin(kx)}{kx}$ . As $x\to 0$ the argument $kx \to 0$ too, so the second factor tends to $1$ by the fundamental limit.∎

Intuition. For tiny angles measured in radians, the sine of the angle is almost exactly the angle itself: $\sin(0.01) = 0.0099998\ldots$ . Geometrically, a short arc of a circle is nearly indistinguishable from the straight chord beneath it, and $\sin x$ is the chord's height while $x$ is the arc's length. The ratio approaches $1$ because the two measurements converge.

Example 1.51 (Using the generalised sine limit). Compute $\lim_{x\to 0} \frac{\sin(5x)}{3x}$ .

Solution.

The expression is not literally $\frac{\sin x}{x}$ , so manufacture that form: multiply and divide by $5$ to match the argument of the sine with its denominator.
$\frac{\sin(5x)}{3x} = \frac{5}{3}\cdot\frac{\sin(5x)}{5x}$ .
As $x\to 0$ we have $5x \to 0$ , so $\frac{\sin(5x)}{5x} \to 1$ .
The limit is $\frac{5}{3}$ , in agreement with the proposition.

□

Example 1.52 (A ratio of two sines). Compute $\lim_{x\to 0}\dfrac{\sin(7x)}{\sin(2x)}$ .

Solution.

Both parts tend to $0$ : indeterminate.
Insert the matching denominators by hand:

\frac{\sin(7x)}{\sin(2x)} = \frac{\sin(7x)}{7x}\cdot\frac{2x}{\sin(2x)}\cdot\frac{7x}{2x}.

The first factor tends to $1$ , the second is the reciprocal of something tending to $1$ and so tends to $1$ , and the third is the constant $\frac72$ .
The limit is $\frac72$ .

□

Example 1.53 (Using an identity first). Compute $\lim_{x\to 0}\dfrac{1 - \cos(4x)}{x\sin(2x)}$ .

Solution.

Form $0/0$ . Use $1 - \cos\theta = 2\sin^{2}(\theta/2)$ with $\theta = 4x$ : the numerator is $2\sin^{2}(2x)$ .
The quotient becomes $\dfrac{2\sin^{2}(2x)}{x\sin(2x)} = \dfrac{2\sin(2x)}{x}$ for $x$ near $0$ with $x \ne 0$ .
Write it as $4\cdot\dfrac{\sin(2x)}{2x}$ , which tends to $4$ .

Sanity check with small numbers: at $x = 0.01$ the original expression is $0.00079989/0.00019999 \approx 3.9997$ , close to $4$ .□

Example 1.54 (A trigonometric limit away from zero). Compute $\lim_{x\to \pi}\dfrac{\sin x}{x - \pi}$ .

Solution.

Substitution gives $0/0$ , but the standard limit is about the argument tending to $0$ , not $\pi$ . Shift: let $u = x - \pi$ , so $u \to 0$ and $x = u + \pi$ .
$\sin(u + \pi) = -\sin u$ , so the quotient is $\dfrac{-\sin u}{u}$ .
This tends to $-1$ .

The substitution $u = x - a$ is the general way to move a limit to the origin where the standard results live.□

Pitfall. Every one of these limits requires radians. In degrees, $\lim_{x\to 0}\frac{\sin x}{x} = \frac{\pi}{180}$ , because $\sin$ of $x$ degrees is $\sin(\pi x/180)$ in radians. A calculator left in degree mode will quietly produce $0.01745$ where you expect $1$ , and every derivative of a trigonometric function in later chapters would acquire the same stray factor.

Exercises for §1.11 →

1.12Continuity

Definition 1.55 (Continuity at a point). A function $f$ is continuous at $x = a$ when all three conditions hold: $f(a)$ is defined; $\lim_{x\to a} f(x)$ exists; and

\lim_{x\to a} f(x) = f(a).

The three conditions are not decoration. The first fails for $\frac{x^{2}-1}{x-1}$ at $x = 1$ , the second for $\lfloor x\rfloor$ at every integer, the third for a function deliberately given the wrong value at one point. Any one failure is a discontinuity, and which condition fails names the type.

Definition 1.56 (One-sided continuity and continuity on an interval). $f$ is continuous from the right at $a$ if $\lim_{x\to a^+} f(x) = f(a)$ , and continuous from the left if $\lim_{x\to a^-} f(x) = f(a)$ . $f$ is continuous on an interval if it is continuous at every point of the interval, where at an endpoint only the appropriate one-sided continuity is required.

The endpoint clause is what allows $f(x) = \sqrt{x}$ to be called continuous on $[0,\infty)$ even though it has no left-hand limit at $0$ : nothing exists to the left, so nothing is demanded.

Now the classification. Four kinds of discontinuity cover everything you will meet.

Removable discontinuity (a hole): $\lim_{x\to a} f(x)$ exists but $f(a)$ is undefined or has the wrong value. Example: $f(x) = \frac{x^{2}-1}{x-1}$ at $x = 1$ , where the limit is $2$ and the value is missing. It is called removable because redefining $f(1) = 2$ repairs it.

Jump discontinuity: both one-sided limits exist but differ. Example: $\lfloor x\rfloor$ at $x = 2$ , with limits $1$ from the left and $2$ from the right. No redefinition of $f(2)$ can help; the two sides disagree, and the gap is real.

Infinite discontinuity: at least one one-sided limit is $\pm\infty$ . Example: $1/x$ at $x = 0$ , or $\ln x$ at $x = 0$ from the right. The graph has a vertical asymptote there.

Oscillating discontinuity: the one-sided limits fail to exist for reasons other than blowing up. Example: $\sin(1/x)$ at $x = 0$ , which takes every value in $[-1,1]$ infinitely often in every neighbourhood of the origin.

Theorem 1.57 (Algebra of continuous functions). If $f$ and $g$ are continuous at $a$ and $c$ is a constant, then $f + g$ , $f - g$ , $cf$ and $fg$ are continuous at $a$ , and so is $f/g$ provided $g(a) \ne 0$ .

Proof. Each statement is a limit law read through the definition of continuity. For the product, $\lim_{x\to a} f(x)g(x) = \left(\lim_{x\to a} f(x)\right)\left(\lim_{x\to a} g(x)\right) = f(a)g(a)$ , which is exactly the assertion that $fg$ is continuous at $a$ . The others are identical, and the proviso $g(a) \ne 0$ in the quotient case is the quotient law's hypothesis, since $\lim_{x\to a} g(x) = g(a)$ .∎

Theorem 1.58 (Continuity of composites). If $\lim_{x\to a} g(x) = b$ and $f$ is continuous at $b$ , then

\lim_{x\to a} f(g(x)) = f\!\left(\lim_{x\to a} g(x)\right) = f(b).

In particular, if $g$ is continuous at $a$ and $f$ is continuous at $g(a)$ , then $f \circ g$ is continuous at $a$ .

Note that the outer function must be continuous, not merely have a limit: the inner function may hit the value $b$ exactly, and then $f(b)$ itself is used.∎

Between them these theorems settle continuity for essentially every formula in a calculus course. Polynomials are continuous on $\mathbb{R}$ ; rational functions on their domains; $\sqrt[n]{x}$ , $\sin x$ , $\cos x$ , $e^x$ , $\ln x$ , and the inverse trigonometric functions on their domains; and any finite combination of these by sums, products, quotients and compositions is continuous wherever it is defined. That is the theorem behind the advice "try substitution first".

Intuition. A continuous function is one you can draw without lifting your pen: no gaps, no jumps, no teleporting.

A hole is a missing stepping stone in an otherwise complete path. A jump is a staircase — you leap to a new height. An infinite discontinuity is a cliff. An oscillating one is a stretch of road that shakes so violently you cannot say where it is heading.

Example 1.59 (Continuity of a piecewise function). Is $f(x) = \begin{cases} x+1, & x < 1 \\ 2, & x=1 \\ x^2, & x > 1 \end{cases}$ continuous at $x=1$ ?

Solution.

Left-hand limit: $\lim_{x\to 1^-}(x+1) = 2$ .
Right-hand limit: $\lim_{x\to 1^+} x^{2} = 1$ .
The one-sided limits differ, so $\lim_{x\to 1} f(x)$ does not exist and the second condition of the definition fails.
$f$ has a jump discontinuity at $x=1$ . The value $f(1) = 2$ matches neither side, but the limit had already decided the question.

□

Example 1.60 (Choosing a constant to make a function continuous). Find the value of $c$ for which

f(x) = \begin{cases} cx^2 + 2x, & x < 2 \\ x^3 - cx, & x \ge 2 \end{cases}

is continuous at $x = 2$ .

Solution.

Each branch is a polynomial, so each one-sided limit is found by substitution.
From the left: $\lim_{x\to 2^-}(cx^{2} + 2x) = 4c + 4$ .
From the right, which also gives the value $f(2)$ : $\lim_{x\to 2^+}(x^{3} - cx) = 8 - 2c$ .
Continuity requires them to be equal: $4c + 4 = 8 - 2c$ , so $6c = 4$ and $c = \tfrac23$ .
Check: both sides then equal $4\cdot\tfrac23 + 4 = \tfrac{20}{3}$ , and $8 - \tfrac43 = \tfrac{20}{3}$ . They agree.

□

Example 1.61 (Classifying every discontinuity of a formula). Locate and classify the discontinuities of $f(x) = \dfrac{x - 2}{x^{2} - 4}$ and of $g(x) = \dfrac{x+1}{x^{2} - 4}$ .

Solution. Both denominators vanish at $x = \pm 2$ , and nowhere else are the functions undefined.

For $f$ : cancelling gives $f(x) = \frac{x-2}{(x-2)(x+2)} = \frac{1}{x+2}$ for $x \ne 2$ . At $x = 2$ the limit is $\frac14$ and the value is undefined — a removable discontinuity, repaired by setting $f(2) = \frac14$ . At $x = -2$ the reduced form still has a zero denominator and nonzero numerator, so the discontinuity is infinite.

For $g$ nothing cancels. At both $x = 2$ and $x = -2$ the numerator is nonzero ( $3$ and $-1$ ), so both discontinuities are infinite, and the graph has two vertical asymptotes.□

Example 1.62 (Continuity through a composition). Where is $h(x) = \ln\!\left(\dfrac{x}{x - 3}\right)$ continuous?

Solution. Write $h = f \circ g$ with $g(x) = \frac{x}{x-3}$ and $f(y) = \ln y$ . The inner function is rational and continuous except at $x = 3$ ; the outer is continuous on $(0,\infty)$ .

By the composition theorem, $h$ is continuous at every $x$ where $g$ is continuous and $g(x) > 0$ . The quotient $\frac{x}{x-3}$ is positive when $x$ and $x-3$ share a sign, that is for $x < 0$ or $x > 3$ .

So $h$ is continuous on $(-\infty, 0) \cup (3, \infty)$ , which is exactly its domain — as the theorem promises for a composition of standard functions.□

Pitfall. Continuity and differentiability are different demands. $f(x) = |x|$ is continuous at $0$ — you can draw it without lifting the pen — but has a corner there and no derivative. The implication runs one way only: differentiable at $a$ forces continuous at $a$ , never the reverse.

Exercises for §1.12 →

1.13The Intermediate Value Theorem

Continuity is worth having because it supports theorems that say something exists without saying how to find it. The first of these is the Intermediate Value Theorem.

Theorem 1.63 (Intermediate Value Theorem). Let $f$ be continuous on the closed interval $[a,b]$ and let $N$ be any number strictly between $f(a)$ and $f(b)$ . Then there exists at least one $c$ in $(a,b)$ with

f(c) = N.

Every hypothesis is doing work. The interval must be closed: on $(0,1]$ the continuous function $1/x$ takes no value at the missing endpoint and the conclusion can fail. The function must be continuous on the whole interval: the floor function on $[0,2]$ never takes the value $\tfrac12$ , although $\lfloor 0\rfloor = 0$ and $\lfloor 2\rfloor = 2$ straddle it. And the theorem promises existence, not uniqueness — a continuous function may hit $N$ many times.

A full proof requires the completeness of the real numbers (one takes $c$ to be the supremum of the set of $x$ in $[a,b]$ with $f(x) < N$ and shows that $f(c) = N$ , using continuity to rule out $f(c) < N$ and $f(c) > N$ ), and that argument belongs to real analysis. The idea is exactly the picture: a curve drawn from height $f(a)$ to height $f(b)$ without lifting the pen must cross every horizontal line in between.

The most common use is the special case $N = 0$ : if $f$ is continuous and $f(a)$ and $f(b)$ have opposite signs, then $f$ has a root in $(a,b)$ . This turns the existence of a solution into a pair of sign checks.

Method 1.64 (Bisection: locating a root to any accuracy). To approximate a root of a continuous $f$ on $[a,b]$ with $f(a)f(b) < 0$ :

Compute the midpoint $m = \tfrac{a+b}{2}$ and the sign of $f(m)$ .
If $f(a)$ and $f(m)$ have opposite signs, the root lies in $[a,m]$ ; replace $b$ by $m$ . Otherwise it lies in $[m,b]$ ; replace $a$ by $m$ .
Repeat. After $n$ steps the root is trapped in an interval of length $(b-a)/2^n$ .

Each step costs one evaluation and buys one bit of accuracy. To get ten decimal places from a starting interval of length $1$ takes about $34$ steps.

Intuition. The Intermediate Value Theorem is common sense made precise: if it was $30^\circ$ F in the morning and $70^\circ$ F in the afternoon, then at some moment it was exactly $50^\circ$ F. Temperature does not teleport.

It is also the reason a wobbly four-legged table can usually be steadied by rotating it: the "gap" under the fourth leg changes continuously as you turn the table, is positive in one orientation and negative a quarter-turn later, so somewhere in between it is zero.

Example 1.65 (Showing an equation has a solution). Show that $x^{3} - x - 1 = 0$ has a root between $1$ and $2$ .

Solution.

Let $f(x) = x^{3} - x - 1$ , a polynomial and therefore continuous on $[1,2]$ .
$f(1) = 1 - 1 - 1 = -1 < 0$ and $f(2) = 8 - 2 - 1 = 5 > 0$ .
The number $N = 0$ lies between $-1$ and $5$ , so the IVT supplies $c$ in $(1,2)$ with $f(c) = 0$ .

The theorem gives no formula for $c$ ; it guarantees that hunting for it is not a waste of time.□

Example 1.66 (Bisection in practice). Locate the root of $f(x) = x^{3} - x - 1$ to within $0.125$ .

Solution. Start with $[1,2]$ , where $f(1) = -1$ and $f(2) = 5$ .

$m = 1.5$ : $f(1.5) = 3.375 - 1.5 - 1 = 0.875 > 0$ . The sign change is now between $1$ and $1.5$ ; take $[1, 1.5]$ .
$m = 1.25$ : $f(1.25) = 1.953125 - 1.25 - 1 = -0.296875 < 0$ . Take $[1.25, 1.5]$ .
$m = 1.375$ : $f(1.375) = 2.5996 - 1.375 - 1 = 0.2246 > 0$ . Take $[1.25, 1.375]$ .

The root lies in $[1.25, 1.375]$ , an interval of length $0.125$ , so $1.3125$ is within $0.0625$ of it. (The true root is $1.3247\ldots$ , the plastic number.)□

Example 1.67 (An intermediate value other than zero). Show that the equation $\cos x = x$ has a solution in $[0,1]$ .

Solution.

Rewrite it as a root problem: let $g(x) = \cos x - x$ , continuous everywhere as a difference of continuous functions.
$g(0) = 1 - 0 = 1 > 0$ and $g(1) = \cos 1 - 1 \approx 0.5403 - 1 = -0.4597 < 0$ .
By the IVT there is $c$ in $(0,1)$ with $g(c) = 0$ , that is $\cos c = c$ .

Sanity check: bisection from here converges to $c \approx 0.739$ , the number your calculator produces if you press cosine repeatedly.□

Pitfall. The converse of the IVT is false. A function that takes every intermediate value need not be continuous: $\sin(1/x)$ extended by $f(0) = 0$ takes every value in $[-1,1]$ on every interval around the origin, yet is discontinuous at $0$ . "Hits every value in between" is a consequence of continuity, not a test for it.

Exercises for §1.13 →

1.14Limits at infinity and horizontal asymptotes

A limit at infinity asks what a function settles down to as $x$ grows without bound. The symbol $\infty$ is not a place $x$ arrives at, so the definition again replaces "close to" with an inequality — this time on the input side.

Definition 1.68 (Limit at infinity). $\lim_{x\to\infty} f(x) = L$ means: for every $\varepsilon > 0$ there is a number $M$ such that

x > M \quad\Longrightarrow\quad |f(x) - L| < \varepsilon.

$\lim_{x\to-\infty} f(x) = L$ is the same with $x < M$ .

Definition 1.69 (Horizontal asymptote). The line $y = L$ is a horizontal asymptote of $f$ if $\lim_{x\to\infty} f(x) = L$ or $\lim_{x\to-\infty} f(x) = L$ .

A graph may have two horizontal asymptotes, one in each direction — $\arctan x$ has $y = \pi/2$ and $y = -\pi/2$ — or one, or none. Nothing forbids the graph from crossing its horizontal asymptote, and $\frac{\sin x}{x}$ crosses $y = 0$ infinitely often; the asymptote describes eventual behaviour, not a barrier.

The basic fact behind every computation is that for any $r > 0$ ,

\lim_{x\to\infty}\frac{1}{x^r} = 0,

which follows straight from the definition: given $\varepsilon$ , take $M = \varepsilon^{-1/r}$ . For a rational function, divide numerator and denominator by the highest power of $x$ in the denominator; every term that still has an $x$ underneath then dies. This gives the familiar trichotomy for $f(x) = p(x)/q(x)$ : if $\deg p < \deg q$ the limit is $0$ ; if $\deg p = \deg q$ it is the ratio of leading coefficients; if $\deg p > \deg q$ the function grows without bound and there is no horizontal asymptote.

Growth rates settle most of the remaining cases. As $x \to \infty$ ,

\ln x \ll x^r \ll e^x

for any $r > 0$ : the logarithm grows more slowly than every positive power, and the exponential faster than every power. Whenever a quotient mixes these families, the faster-growing family decides the limit. In the other direction $\lim_{x\to-\infty}e^x = 0$ , so $y = 0$ is a horizontal asymptote of $e^x$ to the left only.

Radicals need one extra care. For $x > 0$ , $\sqrt{x^{2}} = x$ ; for $x < 0$ , $\sqrt{x^{2}} = -x$ . Dividing by $x$ inside a root as $x \to -\infty$ therefore introduces a sign, and forgetting it is the single most common error in this section.

Intuition. Limits at infinity are about the eventual balance of power. In $\frac{3x^{2} + 100x}{5x^{2} + 1}$ the term $100x$ is far bigger than $3x^{2}$ at $x = 10$ , and the quotient is nowhere near $3/5$ there. But at $x = 10^6$ the squares dwarf everything else. "At infinity" means: wait long enough, and the leading terms win.

Example 1.70 (A rational function and the sign of a radical). Compute $\lim_{x\to\infty}\dfrac{\sqrt{x^{2}+1}}{x - 3}$ and $\lim_{x\to-\infty}\dfrac{\sqrt{x^{2}+1}}{x-3}$ .

Solution. Divide numerator and denominator by $x$ , but carry the root's sign carefully.

As $x \to +\infty$ we may write $x = \sqrt{x^{2}}$ , so

\frac{\sqrt{x^2+1}}{x-3} = \frac{\sqrt{(x^2+1)/x^2}}{(x-3)/x} = \frac{\sqrt{1 + 1/x^2}}{1 - 3/x} \longrightarrow \frac{\sqrt{1}}{1} = 1.

As $x \to -\infty$ the quantity $x$ is negative, so $x = -\sqrt{x^{2}}$ and dividing the numerator by $x$ means dividing the radicand by $x^{2}$ and inserting a minus sign:

\frac{\sqrt{x^2+1}}{x-3} = \frac{-\sqrt{1+1/x^2}}{1 - 3/x} \longrightarrow \frac{-1}{1} = -1.

The graph has two horizontal asymptotes, $y = 1$ to the right and $y = -1$ to the left. Sanity check at $x = -1000$ : the numerator is about $1000$ and the denominator about $-1003$ , giving $\approx -0.997$ .□

Example 1.71 (A difference of radicals). Compute $\lim_{x\to\infty}\left(\sqrt{x^{2} + 3x} - x\right)$ .

Solution.

The form is $\infty - \infty$ : both pieces grow without bound and the difference is undecided.
Multiply and divide by the conjugate $\sqrt{x^{2}+3x}+x$ :

\sqrt{x^2+3x} - x = \frac{(x^2 + 3x) - x^2}{\sqrt{x^2+3x}+x} = \frac{3x}{\sqrt{x^2+3x}+x}.

Divide top and bottom by $x > 0$ : $\dfrac{3}{\sqrt{1 + 3/x} + 1}$ .
As $x \to \infty$ this tends to $\frac{3}{2}$ .

Sanity check at $x = 100$ : $\sqrt{10300} = 101.489$ , and $101.489 - 100 = 1.489$ , close to $1.5$ .□

Example 1.72 (Exponentials at both ends). Find all horizontal asymptotes of $f(x) = \dfrac{e^{2x} - 1}{e^{2x} + 3}$ .

Solution. As $x \to \infty$ , $e^{2x} \to \infty$ and both parts blow up, so divide by $e^{2x}$ :

\frac{1 - e^{-2x}}{1 + 3e^{-2x}} \longrightarrow \frac{1 - 0}{1 + 0} = 1.

As $x \to -\infty$ , $e^{2x} \to 0$ and no work is needed: the quotient tends to $\frac{-1}{3}$ .

So there are two horizontal asymptotes, $y = 1$ and $y = -\tfrac13$ , and the function increases from one to the other. (This is $\tanh$ -like behaviour, and indeed $\frac{e^{2x}-1}{e^{2x}+1} = \tanh x$ .)□

Example 1.73 (Growth rates decide). Compute $\lim_{x\to\infty}\dfrac{(\ln x)^2}{x}$ and $\lim_{x\to\infty}\dfrac{x^{10}}{e^x}$ .

Solution. Both are $\infty/\infty$ . Using the hierarchy $\ln x \ll x^r \ll e^x$ : the numerator $(\ln x)^2$ grows more slowly than any positive power of $x$ , in particular more slowly than $x$ , so the first limit is $0$ . The exponential in the second beats every polynomial, so the second limit is $0$ as well.

Both can be confirmed by L'Hôpital's Rule — the second needs ten applications, which is the honest way to see why the hierarchy is worth memorising.□

Example 1.74 (An oscillation that never settles). Investigate $\lim_{x\to\infty}\sin x$ and $\lim_{x\to\infty}\dfrac{x + \sin x}{x}$ .

Solution. The first does not exist: $\sin x$ returns to $1$ at $x = \frac\pi2 + 2k\pi$ and to $-1$ at $x = -\frac\pi2 + 2k\pi$ for arbitrarily large $k$ , so no single $L$ has all far-out values within $\tfrac12$ of it.

The second does exist. Write it as $1 + \frac{\sin x}{x}$ ; the oscillating part is bounded by $\pm\frac1x$ , which tends to $0$ by the Squeeze Theorem. The limit is $1$ , so $y = 1$ is a horizontal asymptote even though the graph crosses it infinitely often.□

Pitfall. $\sqrt{x^{2}} = |x|$ , not $x$ . When $x \to -\infty$ , pulling an $x$ out of a square root costs a minus sign, and skipping it turns a limit of $-1$ into $+1$ . A quick check: plug in a large negative number and look at the sign of the answer before you believe the algebra.

Exercises for §1.14 →

1.15Slant asymptotes

When a rational function's numerator is exactly one degree higher than its denominator, the graph has no horizontal asymptote — it runs off to infinity — yet it still approaches a line.

Definition 1.75 (Slant (oblique) asymptote). The line $y = mx + b$ with $m \ne 0$ is a slant asymptote of $f$ if

\lim_{x\to\infty}\,[f(x) - (mx + b)] = 0

or the same limit as $x \to -\infty$ .

The definition says the vertical gap between curve and line closes, which is the same standard a horizontal asymptote meets with $m = 0$ . For a rational function the line is found by long division: if

\frac{p(x)}{q(x)} = mx + b + \frac{r(x)}{q(x)}

with $\deg r < \deg q$ , then the remainder term tends to $0$ at infinity and $y = mx + b$ is the asymptote. A rational function has a slant asymptote exactly when $\deg p = \deg q + 1$ , and then the quotient of the division is the asymptote.

Intuition. Divide $18$ by $4$ and you get $4$ remainder $2$ : for rough purposes $18/4$ "is" $4$ . Polynomial long division does the same thing with functions, and the remainder is the part that becomes negligible once $x$ is large. The quotient is the simple function your rational function eventually imitates.

Example 1.76 (Finding a slant asymptote by division). Find the slant asymptote of $f(x) = \dfrac{x^{2} + 1}{x - 1}$ .

Solution.

Degrees are $2$ and $1$ , differing by one, so a slant asymptote exists.
Long division: $x^{2} + 1 = (x-1)(x+1) + 2$ , so

f(x) = x + 1 + \frac{2}{x-1}.

The remainder $\frac{2}{x-1}$ tends to $0$ as $x \to \pm\infty$ , so $y = x + 1$ is a slant asymptote in both directions.
The sign of the remainder also tells you the side: for $x > 1$ the curve lies above the line, for $x < 1$ below it.

The function also has a vertical asymptote at $x = 1$ , where the remainder blows up.□

Example 1.77 (A slant asymptote with a radical). Show that $y = x$ is a slant asymptote of $f(x) = \sqrt{x^{2} + 4x}$ as $x \to \infty$ , and find the asymptote as $x\to-\infty$ .

Solution. For the right-hand direction, consider $\sqrt{x^{2}+4x} - x$ and rationalise:

\sqrt{x^2+4x} - x = \frac{4x}{\sqrt{x^2+4x}+x} = \frac{4}{\sqrt{1 + 4/x} + 1} \longrightarrow 2.

So the gap tends to $2$ , not $0$ : the asymptote is not $y = x$ but $y = x + 2$ . Indeed $\lim_{x\to\infty}[f(x) - (x+2)] = 2 - 2 = 0$ .

As $x \to -\infty$ the same computation with $x = -\sqrt{x^{2}}$ gives $\sqrt{x^{2}+4x} + x \to -2$ , so $f(x) - (-x - 2) \to 0$ and the left-hand asymptote is $y = -x-2$ . The graph is a hyperbola-like curve with two slant asymptotes meeting at $(-2,0)$ .

Moral: compute the limit of $f(x) - (mx+b)$ rather than guessing the line from the leading term.□

Example 1.78 (When the difference of degrees is two). Does $f(x) = \dfrac{x^{3}}{x - 2}$ have a slant asymptote?

Solution. Degrees differ by two, so the answer should be no, and division confirms it:

\frac{x^3}{x-2} = x^2 + 2x + 4 + \frac{8}{x-2}.

The curve approaches the parabola $y = x^{2} + 2x + 4$ , not a line: the difference $f(x) - (x^{2}+2x+4) = \frac{8}{x-2}$ tends to $0$ . This is a curvilinear asymptote, a perfectly respectable object but not covered by the definition of a slant asymptote, which requires a line.□

Pitfall. Long division must be carried out before you read off the asymptote; the ratio of leading terms is not enough. For $\frac{2x^{2} + 5x}{x + 1}$ , the leading behaviour suggests $y = 2x$ , but division gives $2x + 3 - \frac{3}{x+1}$ , so the asymptote is $y = 2x + 3$ . The constant term matters.

Exercises for §1.15 →

1.16L'Hôpital's Rule

Algebra clears an indeterminate form when the cancellation is visible. For $\frac{\ln x}{x-1}$ at $x = 1$ , or $\frac{e^x - 1 - x}{x^{2}}$ at $x = 0$ , nothing factors. L'Hôpital's Rule replaces the values, which say nothing, by the rates of change, which say everything.

Theorem 1.79 (L'Hôpital's Rule). Suppose $f$ and $g$ are differentiable on an open interval containing $a$ (except possibly at $a$ ) and $g'(x) \neq 0$ there. Suppose further that

\lim_{x\to a} f(x) = \lim_{x\to a} g(x) = 0 \qquad\text{or}\qquad \lim_{x\to a}|f(x)| = \lim_{x\to a}|g(x)| = \infty.

Then

\lim_{x\to a} \frac{f(x)}{g(x)} = \lim_{x\to a} \frac{f'(x)}{g'(x)},

provided the limit on the right exists or is $\pm\infty$ . The same holds for one-sided limits and for $a = \pm\infty$ .

Proof. The idea, in the $0/0$ case with $f(a) = g(a) = 0$ and $f', g'$ continuous at $a$ with $g'(a) \ne 0$ :

\frac{f(x)}{g(x)} = \frac{f(x) - f(a)}{g(x) - g(a)} = \frac{\dfrac{f(x)-f(a)}{x-a}}{\dfrac{g(x)-g(a)}{x-a}} \longrightarrow \frac{f'(a)}{g'(a)}.

The general statement needs the Cauchy Mean Value Theorem: if $f$ and $g$ are continuous on $[a,x]$ and differentiable on $(a,x)$ , there is a point $\xi$ between $a$ and $x$ with

[f(x) - f(a)]\,g'(\xi) = [g(x) - g(a)]\,f'(\xi).

With $f(a) = g(a) = 0$ this rearranges to $\frac{f(x)}{g(x)} = \frac{f'(\xi)}{g'(\xi)}$ . As $x \to a$ the intermediate point $\xi$ is squeezed to $a$ as well, so the left side inherits the limit of $f'/g'$ . The $\infty/\infty$ case is proved by a similar but more delicate argument, which belongs to real analysis.∎

Three features of the statement deserve emphasis.

It is not the quotient rule. You differentiate the numerator and the denominator separately and form a new quotient; applying the quotient rule here is the single most common way to get a wrong answer.

It is not a chain-rule trick or a cancellation of the $x$ 's. The reason it works is the displayed computation above: near $a$ , both functions are approximately linear, $f(x) \approx f'(a)(x-a)$ and $g(x) \approx g'(a)(x-a)$ , and the common factor $(x-a)$ cancels. The rule is that approximation made exact.

The implication is one-way. If $\lim f'/g'$ exists, it equals $\lim f/g$ . If $\lim f'/g'$ fails to exist, nothing follows — the original limit may exist anyway, and you must use another method.

Method 1.80 (Handling the seven indeterminate forms).

$\frac{0}{0}$ and $\frac{\infty}{\infty}$ : apply the rule directly, and re-check the form after each application.
$0 \cdot \infty$ : turn the product $fg$ into $\dfrac{f}{1/g}$ or $\dfrac{g}{1/f}$ , whichever differentiates more kindly.
$\infty - \infty$ : combine over a common denominator, rationalise, or factor out the dominant term, to reach a quotient.
$0^0$ , $1^\infty$ , $\infty^{0}$ : set $y = f(x)^{g(x)}$ , take logarithms to get $\ln y = g(x)\ln f(x)$ , which is a $0\cdot\infty$ form; find $\lim \ln y = L$ and conclude $\lim y = e^{L}$ .

The last step of the logarithm method uses continuity of the exponential: if $\ln y \to L$ then $y = e^{\ln y} \to e^L$ by the composition theorem. Forgetting to exponentiate at the end — reporting $L$ instead of $e^L$ — is the classic slip.

Intuition. Both functions are heading to zero, so their values tell you nothing. The derivative measures how fast each one gets there, and the ratio of speeds is exactly the information the ratio of values lost.

If one application still gives $\frac00$ , the first-order speeds cancelled too. Applying the rule again compares accelerations, and so on, until the rates finally differ enough to settle the limit.

Example 1.81 (A single application). Compute $\lim_{x\to 0} \frac{e^x - 1}{\sin x}$ .

Solution.

Substitution gives $\frac{e^{0}-1}{\sin 0} = \frac{0}{0}$ , so the rule applies.
Differentiate separately: $f'(x) = e^x$ and $g'(x) = \cos x$ .
The new limit is $\lim_{x\to 0}\frac{e^x}{\cos x} = \frac{1}{1} = 1$ , no longer indeterminate.

□

Example 1.82 (Repeated applications). Compute $\lim_{x\to 0} \frac{x - \sin x}{x^{3}}$ .

Solution.

Substitution gives $\frac{0}{0}$ ; apply the rule to get $\lim_{x\to 0}\frac{1-\cos x}{3x^{2}}$ .
Still $\frac{0}{0}$ ; apply it again to get $\lim_{x\to 0}\frac{\sin x}{6x}$ .
Still $\frac{0}{0}$ ; a third application gives $\lim_{x\to 0}\frac{\cos x}{6}$ .
This is a definite value, so the limit is $\frac{1}{6}$ .

Each pass peeled away one order of cancellation between $x - \sin x$ and $x^{3}$ .□

Example 1.83 (Converting a 0 · ∞ form). Compute $\lim_{x\to 0^+} x \ln x$ .

Solution.

As $x\to 0^+$ , $x \to 0$ while $\ln x \to -\infty$ : a product, not a quotient, so the rule does not apply yet.
Rewrite $x\ln x = \frac{\ln x}{1/x}$ , now of the form $\frac{-\infty}{\infty}$ .
Differentiate: $\frac{d}{dx}\ln x = \frac{1}{x}$ and $\frac{d}{dx}\frac{1}{x} = -\frac{1}{x^{2}}$ .
So the limit equals $\lim_{x\to 0^+} \frac{1/x}{-1/x^{2}} = \lim_{x\to 0^+}(-x) = 0$ .

The other rewrite, $\frac{x}{1/\ln x}$ , is legal but leads to a messier derivative — choose the arrangement that simplifies.□

Example 1.84 (An ∞ − ∞ form). Compute $\lim_{x\to 0^+}\left(\dfrac{1}{x} - \dfrac{1}{\sin x}\right)$ .

Solution.

Both terms blow up: form $\infty - \infty$ .
Common denominator: $\dfrac{\sin x - x}{x\sin x}$ , now $\frac00$ .
L'Hôpital once: $\dfrac{\cos x - 1}{\sin x + x\cos x}$ , still $\frac00$ .
Again: $\dfrac{-\sin x}{2\cos x - x\sin x} \longrightarrow \dfrac{0}{2} = 0$ .

The limit is $0$ : near the origin $\sin x$ and $x$ agree closely enough that their reciprocals' difference vanishes.□

Example 1.85 (A one-to-the-infinity form via logarithms). Compute $\lim_{x\to\infty}\left(1 + \dfrac{5}{x}\right)^{x}$ .

Solution.

The base tends to $1$ and the exponent to $\infty$ : the form $1^\infty$ , which is indeterminate — a base slightly above $1$ raised to a huge power can do anything.
Let $y = \left(1 + \frac5x\right)^x$ , so $\ln y = x\ln\!\left(1 + \frac5x\right)$ , of form $\infty \cdot 0$ .
Rewrite as a quotient: $\ln y = \dfrac{\ln(1 + 5/x)}{1/x}$ , form $\frac00$ .
Differentiate top and bottom with respect to $x$ . The top has derivative $\dfrac{1}{1+5/x}\cdot\left(-\dfrac{5}{x^{2}}\right)$ and the bottom $-\dfrac{1}{x^{2}}$ , so the quotient is $\dfrac{5}{1 + 5/x} \longrightarrow 5$ .
Hence $\ln y \to 5$ , and by continuity of $\exp$ , $y \to e^{5}$ .

Sanity check: this is the compound-interest limit, and $e^{5} \approx 148.41$ ; at $x = 1000$ the expression is $(1.005)^{1000} \approx 146.58$ , closing in from below.□

Example 1.86 (A zero-to-the-zero form). Compute $\lim_{x\to 0^+} x^{\sin x}$ .

Solution.

Base $\to 0$ , exponent $\to 0$ : form $0^0$ , indeterminate.
Let $y = x^{\sin x}$ , so $\ln y = \sin x\,\ln x$ , of form $0\cdot(-\infty)$ .
Rewrite as $\dfrac{\ln x}{1/\sin x} = \dfrac{\ln x}{\csc x}$ , form $\frac{-\infty}{\infty}$ . L'Hôpital gives

\frac{1/x}{-\csc x\cot x} = -\frac{\sin x\tan x}{x} = -\frac{\sin x}{x}\cdot\tan x \longrightarrow -1 \cdot 0 = 0.

So $\ln y \to 0$ and $y \to e^{0} = 1$ .

□

Example 1.87 (An infinity-to-the-zero form). Compute $\lim_{x\to\infty} x^{1/x}$ .

Solution.

Base $\to\infty$ , exponent $\to 0$ : form $\infty^{0}$ .
$\ln y = \dfrac{\ln x}{x}$ , of form $\frac{\infty}{\infty}$ .
L'Hôpital: $\dfrac{1/x}{1} = \dfrac1x \to 0$ .
So $y \to e^{0} = 1$ . Sanity check: $1000^{1/1000} = e^{6.9/1000} \approx 1.0069$ .

□

Example 1.88 (Where the rule loops forever). Try to compute $\lim_{x\to\infty}\dfrac{\sqrt{x^{2}+1}}{x}$ with L'Hôpital's Rule.

Solution. The form is $\frac{\infty}{\infty}$ , so the rule applies. Differentiating gives

\frac{x/\sqrt{x^2+1}}{1} = \frac{x}{\sqrt{x^2+1}},

which is the reciprocal of the original expression — still $\frac{\infty}{\infty}$ . A second application returns the original quotient, and the process cycles forever.

The rule was applicable and gave no information, which is allowed: it promises equality of limits, not progress. Algebra settles it at once: $\frac{\sqrt{x^{2}+1}}{x} = \sqrt{1 + 1/x^{2}} \to 1$ .□

Example 1.89 (Where the rule's hypothesis fails). Evaluate $\lim_{x\to 0}\dfrac{x + \cos x}{2x + 1}$ , and say what L'Hôpital's Rule would have given.

Solution. Substitution gives $\frac{0 + 1}{0 + 1} = 1$ . The form is not indeterminate, so the rule does not apply and the limit is simply $1$ .

Applying it anyway gives $\lim_{x\to 0}\frac{1 - \sin x}{2} = \frac12$ , which is wrong. Checking the form before differentiating is not a formality; it is the hypothesis of the theorem.□

Pitfall. Two failure modes to watch for. First, the rule can be applicable and useless: $\lim_{x\to\infty}\frac{x + \sin x}{x}$ has form $\frac\infty\infty$ , and differentiating gives $\frac{1 + \cos x}{1}$ , which has no limit — yet the original limit is $1$ by the Squeeze Theorem. A nonexistent limit for $f'/g'$ tells you nothing about $f/g$ . Second, the answer to a $1^\infty$ , $0^0$ or $\infty^{0}$ problem is $e^L$ , where $L$ is the limit of the logarithm; stopping at $L$ answers a different question.

Summary. $\lim_{x\to a} f(x) = L$ means: for every $\varepsilon > 0$ there is a $\delta > 0$ with $0 < |x-a| < \delta \Rightarrow |f(x)-L| < \varepsilon$ . The value $f(a)$ is irrelevant, the limit is unique, and it exists exactly when both one-sided limits exist and agree. When $\lim f = L$ and $\lim g = M$ both exist, the limit laws give sums, constant multiples, products, powers and roots, and $L/M$ provided $M \ne 0$ ; hence $\lim_{x\to a} p(x) = p(a)$ for a polynomial and $f(a)$ for a rational $f = p/q$ with $q(a) \ne 0$ . Indeterminate forms yield to algebra, to the Squeeze Theorem ( $g \le f \le h$ near $a$ with $\lim g = \lim h = L$ forces $\lim f = L$ ), or to the radian limits $\lim_{x\to 0}\frac{\sin x}{x} = 1$ and $\lim_{x\to 0}\frac{1-\cos x}{x^{2}} = \frac12$ . Continuity at $a$ needs all three of $f(a)$ defined, the limit existing, and $\lim_{x\to a} f(x) = f(a)$ ; it survives sums, products, quotients with $g(a)\ne 0$ , and composition. For $f$ continuous on the closed interval $[a,b]$ , the Intermediate Value Theorem gives $c$ in $(a,b)$ with $f(c) = N$ for every $N$ strictly between $f(a)$ and $f(b)$ . L'Hôpital's Rule replaces $\lim f/g$ by $\lim f'/g'$ only for $\frac00$ and $\frac\infty\infty$ , with $f,g$ differentiable near $a$ , $g' \ne 0$ , and the new limit existing or infinite.

Exercises for §1.16 →

Common Mistakes

Plugging in the value too early before simplifying an indeterminate form. Always check for $0/0$ or $\infty/\infty$ first.
Concluding the limit does not exist just because $f(a)$ is undefined. The function not being defined at $a$ says nothing about the limit.
Trusting a table of values. Sampling can miss oscillation, as with $\sin(\pi/x)$ near $0$ , and rounding error corrupts tables built from very small inputs.
Ignoring one-sided limits for piecewise functions and absolute values. You must check both sides, and the value assigned at the breakpoint decides continuity, never the limit.
Confusing continuity with differentiability. Continuity means no jumps or holes; differentiability means no sharp corners either. $|x|$ is continuous at $0$ but not differentiable.
Forgetting to cancel common factors before declaring a vertical asymptote. A common factor means a hole, not an asymptote.
Writing $\sqrt{x^{2}} = x$ when $x$ is negative. It equals $|x|$ , and the missing minus sign flips the horizontal asymptote as $x \to -\infty$ .
Reading a slant asymptote off the leading terms instead of doing the long division. $\frac{2x^{2}+5x}{x+1}$ approaches $y = 2x + 3$ , not $y = 2x$ .
Using L'Hôpital's Rule when the form is not indeterminate. It only applies to $0/0$ or $\infty/\infty$ .
Applying the quotient rule instead of differentiating numerator and denominator separately in L'Hôpital's Rule.
Stopping at the limit of the logarithm in a $1^\infty$ , $0^0$ or $\infty^{0}$ problem. The answer is $e$ to that power.
Concluding that a limit does not exist because $\lim f'/g'$ does not exist. L'Hôpital's implication runs one way only.
In an $\varepsilon$ – $\delta$ proof, choosing $\delta$ after seeing $x$ , or presenting the backwards analysis as the proof. The proof is the forward verification with a $\delta$ that depends only on $\varepsilon$ .
Applying a limit law whose hypotheses fail: the quotient law needs a nonzero limit in the denominator, and every law needs the individual limits to exist.
Forgetting that $\lim_{x\to 0}\frac{\sin x}{x} = 1$ requires radians. In degrees the value is $\pi/180$ .
Thinking all discontinuities are 'bad' — removable ones are often easy to fix, and the function can be made continuous with a redefinition.

About the practice questions. They check that you can carry out this chapter's computations correctly, and each one is graded on a single answer. They are not proof exercises: working through them confirms the mechanics, not that you could prove the results yourself. For that, re-read the statements above and try to reconstruct their proofs with the page closed.

Derivatives →

Helpful?