Limits & Continuity

(a) Factor and cancel. Substitution gives $\frac{0}{0}$. Factoring both parts,

$$\frac{x^2+x-2}{x^2-4}=\frac{(x+2)(x-1)}{(x+2)(x-2)}=\frac{x-1}{x-2}(x\ne -2).$$

The cancellation is licensed because a limit never inspects $x=-2$ itself. Substituting now gives $\frac{-3}{-4}=\frac{3}{4}$.

(b) Rationalise. Again $\frac{0}{0}$, and the obstacle is the root, so multiply above and below by the conjugate $\sqrt{9+h}+3$:

$$\frac{\sqrt{9+h}-3}{h}=\frac{(9+h)-9}{h\left(\sqrt{9+h}+3\right)}=\frac{1}{\sqrt{9+h}+3}(h\ne 0),$$

which tends to $\frac{1}{3+3}=\frac{1}{6}$.

(c) Combine the fractions, then rationalise. Each term blows up, so the form is $\infty -\infty$ and nothing may be concluded from the pieces. Over the common denominator $t\sqrt{1+t}$,

$$\frac{1}{t\sqrt{1+t}}-\frac{1}{t}=\frac{1-\sqrt{1+t}}{t\sqrt{1+t}}.$$

Now multiply above and below by $1+\sqrt{1+t}$; the numerator becomes $1-(1+t)=-t$, so for $t\ne 0$ the expression equals

$$\frac{-t}{t\sqrt{1+t}\left(1+\sqrt{1+t}\right)}=\frac{-1}{\sqrt{1+t}\left(1+\sqrt{1+t}\right)}⟶\frac{-1}{1\cdot 2}=-\frac{1}{2}.$$

Sanity check: at $t=0.0001$ the original expression is $-0.4999625$.

(a) To the left of $3$ only the first branch is in play, and there $x\ne 3$, so the common factor may be cancelled:

$$\frac{x^2-x-6}{x-3}=\frac{(x-3)(x+2)}{x-3}=x+2⟶5\text{as }x\to 3^{-}.$$

To the right, $\frac{b}{x}+1$ is a rational function whose denominator does not vanish at $3$, so direct substitution applies: ${\lim }_{x\to 3^{+}}f(x)=\frac{b}{3}+1$.

(b) The definition asks for three things. $f(3)$ is defined — it equals $a$, whatever $a$ is. The two-sided limit exists exactly when the one-sided limits agree, that is when $\frac{b}{3}+1=5$, so $b=12$; the common value is then $5$. Finally the limit must equal the value, so $a=5$. The pair is $(a,b)=(5,12)$, and it is unique because each condition determined one constant.

(c) With $b=3$ the right-hand limit is $\frac{3}{3}+1=2$, while the left-hand limit is still $5$. Both one-sided limits exist and they differ, so this is a jump discontinuity, of size $5-2=3$.

The constant $a$ only sets the single value $f(3)$. The failure here is in the second condition — the two-sided limit does not exist at all — and that condition mentions no value of $f$ at $3$. So changing $a$ cannot help: a jump is a disagreement between the two sides, not a mismatch between a side and a point.

(a) $x^2-x-2=(x-2)(x+1)$ and $x^2-3x+2=(x-1)(x-2)$, so

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

The test is not "denominator zero" but "denominator zero after cancelling, with the numerator then nonzero". At $x=2$ the factor cancels, so that is a hole. At $x=1$ the reduced numerator is $2\ne 0$ while the reduced denominator vanishes, so $x=1$ is a vertical asymptote.

(b) Work with the reduced form throughout; it agrees with $f$ everywhere except at $x=2$, which no limit can see.

As $x\to 1^{-}$ the numerator tends to $2>0$ and $x-1\to 0^{-}$, so $f(x)\to -\infty$. As $x\to 1^{+}$ the denominator is small and positive, so $f(x)\to +\infty$.

At $x=2$ the reduced form is continuous, so ${\lim }_{x\to 2}f(x)=\frac{3}{1}=3$ — the limit exists even though $f(2)$ does not, which is exactly what a hole is.

Dividing by $x$, $\frac{x+1}{x-1}=\frac{1+1/x}{1-1/x}\to 1$ in both directions, so ${\lim }_{x\to \pm \infty }f(x)=1$.

(c) Intercepts: $f(x)=0$ only where the reduced numerator vanishes, at $x=-1$; note $x=2$ is not an intercept, it is the hole. The $y$-intercept is $f(0)=\frac{1}{-1}=-1$.

For the sides, rewrite once more:

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

The remainder $\frac{2}{x-1}$ is negative for $x<1$ and positive for $x>1$, so the left branch approaches $y=1$ from below and the right branch from above.

The sketch therefore has two branches. On $(-\infty ,1)$ the curve comes in just under $y=1$ (at $x=-3$ it is $0.5$), falls through $(-1,0)$ and $(0,-1)$, and plunges to $-\infty$ at the wall $x=1$. On $(1,\infty )$ it descends from $+\infty$, passes through the open circle at $(2,3)$, then $(3,2)$ and $(5,1.5)$, settling onto $y=1$ from above. Draw the vertical line $x=1$ and the horizontal line $y=1$ dashed, and the hole as an unfilled circle.

(a) There are three distinct flaws.

Step 1 checks only the form. L'Hôpital's Rule also requires $g^{\prime }(x)\ne 0$ on a punctured neighbourhood of the point. Here $g(x)=x+\cos x$ and $g^{\prime }(x)=1-\sin x$, which vanishes at every $x=\frac{\pi }{2}+2k\pi$ — that is, arbitrarily far out. The hypothesis fails, so the rule may not be invoked at all, and Step 2 is not a legal deduction.

Step 3 is a non sequitur. Bounded does not mean convergent. The quotient $\frac{1+\cos x}{1-\sin x}$ has no limit as $x\to \infty$: it equals $0$ whenever $\cos x=-1$, and it is unbounded near the zeros of $1-\sin x$. So the expression in Step 2 is not bounded either, and in any case a bound is never a value.

The implication runs one way. Even if the limit in Step 2 had merely failed to exist, nothing would follow. L'Hôpital's Rule says: if $\lim f^{\prime }/g^{\prime }$ exists, then it equals $\lim f/g$. A nonexistent $\lim f^{\prime }/g^{\prime }$ tells you nothing about $\lim f/g$. The student's "$1$" was asserted, not derived.

(b) Divide numerator and denominator by $x$, which is legal for $x>0$:

$$\frac{x+\sin x}{x+\cos x}=\frac{1+\frac{\sin x}{x}}{1+\frac{\cos x}{x}}.$$

For $x>0$ we have $-\frac{1}{x}\le \frac{\sin x}{x}\le \frac{1}{x}$, and both bounds tend to $0$, so $\frac{\sin x}{x}\to 0$ by the Squeeze Theorem; the same argument gives $\frac{\cos x}{x}\to 0$. The denominator therefore tends to $1\ne 0$, the quotient law applies, and the limit is

$$\frac{1+0}{1+0}=1.$$

Sanity check: at $x=10^6$ the quotient is $0.9999987$.

(a) Here $x\to -\infty$, so $x<0$ and $\sqrt{x^2}=|x|=-x$. (The radicand is positive for $x<-\frac{3}{4}$, so the expression is defined out there.) Pulling $x^2$ out of the root and $x$ out of the denominator,

$$\frac{\sqrt{4x^2+3x}}{2x+1}=\frac{|x|\sqrt{4+\frac{3}{x}}}{x\left(2+\frac{1}{x}\right)}=\frac{-x\sqrt{4+\frac{3}{x}}}{x\left(2+\frac{1}{x}\right)}=\frac{-\sqrt{4+\frac{3}{x}}}{2+\frac{1}{x}}⟶\frac{-2}{2}=-1.$$

Sanity check: at $x=-10^6$ the quotient is $-1.0000001$. Dropping the minus sign would have produced $+1$, the single most common error in this section.

(b) The form is $\infty -\infty$. Multiply and divide by the conjugate:

$$\sqrt{x^2+5x}-\sqrt{x^2+x}=\frac{(x^2+5x)-(x^2+x)}{\sqrt{x^2+5x}+\sqrt{x^2+x}}=\frac{4x}{\sqrt{x^2+5x}+\sqrt{x^2+x}}.$$

Now $x\to +\infty$, so $x>0$ and $\sqrt{x^2}=x$; dividing top and bottom by $x$ gives

$$\frac{4}{\sqrt{1+\frac{5}{x}}+\sqrt{1+\frac{1}{x}}}⟶\frac{4}{1+1}=2.$$

(c) The form is $\infty \cdot 0$. Rationalise the bracket:

$$x\left(\sqrt{1+\frac{2}{x}}-1\right)=x\cdot \frac{\left(1+\frac{2}{x}\right)-1}{\sqrt{1+\frac{2}{x}}+1}=\frac{x\cdot \frac{2}{x}}{\sqrt{1+\frac{2}{x}}+1}=\frac{2}{\sqrt{1+\frac{2}{x}}+1},$$

which tends to $\frac{2}{1+1}=1$.

(a) Preliminary analysis (scratch work, not the proof). For $x\ne 0$,

$$\left|\frac{1}{x}-\frac{1}{2}\right|=\left|\frac{2-x}{2x}\right|=\frac{|x-2|}{2|x|}.$$

The factor $|x-2|$ is under our control; the factor $\frac{1}{2|x|}$ is not, and it is unbounded as $x$ nears $0$. So tame it first: agree to keep $|x-2|<1$. Then $1<x<3$, hence $|x|>1$ and $\frac{1}{2|x|}<\frac{1}{2}$. Under that agreement the whole expression is below $\frac{|x-2|}{2}$, which is below $\epsilon$ as soon as $|x-2|<2\epsilon$.

Proof. Let $\epsilon >0$ be given and set

$$\delta =\min \left(1,2\epsilon \right),$$

a positive number depending on $\epsilon$ alone. Suppose $0<|x-2|<\delta$. From $|x-2|<1$ we get $x>1$, so $x\ne 0$ and $2|x|>2$. Therefore

$$\left|\frac{1}{x}-\frac{1}{2}\right|=\frac{|x-2|}{2|x|}<\frac{|x-2|}{2}<\frac{\delta }{2}\le \epsilon .$$

Since $\epsilon >0$ was arbitrary, ${\lim }_{x\to 2}\frac{1}{x}=\frac{1}{2}$. The $\min$ is the standard device: one bound tames the awkward factor, the other delivers the tolerance.

(b) Suppose, for contradiction, that some real $L$ satisfied the definition, and take $\epsilon =1$. Let $\delta >0$ be arbitrary and put

$$x=\min \left(\frac{\delta }{2},\frac{1}{|L|+2}\right)>0.$$

Then $0<|x-0|=x<\delta$, so the definition would force $\left|\frac{1}{x}-L\right|<1$. But $x\le \frac{1}{|L|+2}$ gives $\frac{1}{x}\ge |L|+2$, and by the reverse triangle inequality

$$\left|\frac{1}{x}-L\right|\ge \frac{1}{x}-|L|\ge (|L|+2)-|L|=2>1,$$

a contradiction. No $\delta$ works for $\epsilon =1$, whatever $L$ is, so the limit does not exist.

The computation also shows how it fails: $\frac{1}{x}$ exceeds every bound as $x\to 0^{+}$, which is what the one-sided statement ${\lim }_{x\to 0^{+}}\frac{1}{x}=\infty$ records.

(a) Manufacture the standard forms by matching each argument with its own denominator:

$$\frac{\sin (3x)\tan (5x)}{x^2}=3\cdot \frac{\sin (3x)}{3x}\cdot 5\cdot \frac{\tan (5x)}{5x}.$$

As $x\to 0$ both $3x\to 0$ and $5x\to 0$, so each quotient tends to $1$ (the second because $\frac{\tan u}{u}=\frac{\sin u}{u}\cdot \frac{1}{\cos u}\to 1\cdot 1$). The limit is $3\cdot 5=15$.

(b) Divide numerator and denominator by $x^2$, so that both acquire limits and the quotient law becomes available:

$$\frac{1-\cos (3x)}{x\sin (4x)}=\frac{\frac{1-\cos (3x)}{x^2}}{\frac{\sin (4x)}{x}}.$$

The numerator is $9\cdot \frac{1-\cos (3x)}{(3x{)}^2}\to 9\cdot \frac{1}{2}=\frac{9}{2}$, and the denominator is $4\cdot \frac{\sin (4x)}{4x}\to 4$. Since $4\ne 0$ the quotient law applies and the limit is $\frac{9/2}{4}=\frac{9}{8}$.

(c) Factor $\sin x$ out of the numerator, which is what makes the cancellation visible:

$$\tan x-\sin x=\sin x\left(\frac{1}{\cos x}-1\right)=\frac{\sin x(1-\cos x)}{\cos x}.$$

Hence, for $x\ne 0$ near $0$,

$$\frac{\tan x-\sin x}{x^3}=\frac{\sin x}{x}\cdot \frac{1-\cos x}{x^2}\cdot \frac{1}{\cos x}⟶1\cdot \frac{1}{2}\cdot 1=\frac{1}{2}.$$

Sanity check: at $x=0.001$ the expression is $0.5000001$.

(a) Both bounds are polynomials, hence continuous, so their limits at $1$ are found by substitution: $3x\to 3$ and $x^3+2\to 3$. The two bounds have the same limit, so by the Squeeze Theorem ${\lim }_{x\to 1}f(x)=3$.

Nothing else about $f$ is needed, and that is the point of the theorem: the hypothesis traps $f$ between two functions which are forced together at $x=1$, so $f$ has no room left to do anything else. (Note the value $f(1)$ is still unknown and irrelevant.)

(b) By hypothesis there are ${\delta }_0>0$ and $K>0$ with $|g(x)|\le K$ whenever $0<|x-a|<{\delta }_0$. (If the given bound happens to be $0$, replace it by any positive number; we need $K>0$ only in order to divide by it.)

Let $\epsilon >0$. Since ${\lim }_{x\to a}f(x)=0$, there is ${\delta }_1>0$ such that

$$0<|x-a|<{\delta }_1⟹|f(x)-0|<\frac{\epsilon }{K}.$$

Put $\delta =\min ({\delta }_0,{\delta }_1)>0$. If $0<|x-a|<\delta$ then both estimates are available, so

$$|f(x)g(x)-0|=|f(x)||g(x)|\le |f(x)|K<\frac{\epsilon }{K}\cdot K=\epsilon .$$

As $\epsilon >0$ was arbitrary, ${\lim }_{x\to a}f(x)g(x)=0$. $■$

(c) Take $f(x)=\sqrt[3]{x}$ and $g(x)=\sin (1/x)$, with $a=0$. Then $f$ is continuous at $0$ with $f(0)=0$, so ${\lim }_{x\to 0}f(x)=0$; and $|g(x)|\le 1$ on every punctured neighbourhood of $0$. By (b) the limit is $0$.

The product law is unavailable because it requires both factors to have limits, and ${\lim }_{x\to 0}\sin (1/x)$ does not exist: at $x=\frac{2}{(4k+1)\pi }$ we get $\sin (1/x)=1$, and at $x=\frac{2}{(4k+3)\pi }$ we get $\sin (1/x)=-1$, and both families of points reach arbitrarily close to $0$. No single $L$ can be approached. The boundedness of $g$ is all that survives — and (b) says that is enough.

(a) Turn it into a root problem: let $h(x)=e^x+x^2-3$, so the solutions of the equation are exactly the zeros of $h$. As a sum of continuous functions, $h$ is continuous on every closed interval.

$$h(-2)=e^{-2}+4-3=e^{-2}+1\approx 1.135>0,h(0)=1+0-3=-2<0,h(1)=e-2\approx 0.718>0.$$

Since $h(-2)>0>h(0)$ and $h$ is continuous on $[-2,0]$, the IVT (with $N=0$) gives a zero in $(-2,0)$. Since $h(0)<0<h(1)$, the IVT on $[0,1]$ gives a zero in $(0,1)$. The two intervals are disjoint, so the zeros are distinct and the equation has at least two solutions.

To halve each interval, evaluate at the midpoints: $h(-1)=e^{-1}-2\approx -1.632<0$, so the first root lies in $(-2,-1)$; and $h(0.5)=e^{0.5}-2.75\approx -1.101<0$, so the second lies in $(0.5,1)$. Both intervals have length $1$ or less, as required.

(b) Define $g(x)=f(x)-x$ on $[0,1]$. It is continuous, being a difference of continuous functions.

Because $f$ takes values in $[0,1]$ we have $f(0)\ge 0$ and $f(1)\le 1$, so

$$g(0)=f(0)-0\ge 0,g(1)=f(1)-1\le 0.$$

If $g(0)=0$ then $c=0$ is a fixed point; if $g(1)=0$ then $c=1$ is. Otherwise $g(0)>0>g(1)$, so $0$ lies strictly between $g(1)$ and $g(0)$, and the IVT applied to $g$ on the closed interval $[0,1]$ supplies $c\in (0,1)$ with $g(c)=0$, that is $f(c)=c$. $■$

(c) Take

$$f(x)=\{\begin{array}{ll}1,& 0\le x\le \frac{1}{2},\\ 0,& \frac{1}{2}<x\le 1.\end{array}$$

Its values are $0$ and $1$, both in $[0,1]$, so it is a map $[0,1]\to [0,1]$. If $x\le \frac{1}{2}$ then $f(x)=1\ne x$, since $x\le \frac{1}{2}<1$; if $x>\frac{1}{2}$ then $f(x)=0\ne x$, since $x>0$. So $f$ has no fixed point.

The hypothesis violated is continuity: $f$ has a jump at $x=\frac{1}{2}$. One sees exactly where the proof of (b) breaks: $g(\frac{1}{2})=1-\frac{1}{2}=\frac{1}{2}>0$, while $g(x)=-x<0$ for every $x$ just to the right of $\frac{1}{2}$, so $g$ steps over the value $0$ without ever taking it — which is precisely what the IVT forbids a continuous function to do.

(a) Dividing $2x^2-3x+1$ by $x-2$: the first quotient term is $2x$, since $2x(x-2)=2x^2-4x$, and subtracting leaves $x+1$; the next term is $1$, since $1\cdot (x-2)=x-2$, and subtracting leaves the remainder $3$. Hence

$$f(x)=2x+1+\frac{3}{x-2}.$$

Check by multiplying back: $(2x+1)(x-2)+3=2x^2-3x-2+3=2x^2-3x+1$. ✓

Now $f(x)-(2x+1)=\frac{3}{x-2}\to 0$ as $x\to +\infty$ and as $x\to -\infty$, which is exactly the definition of a slant asymptote. So $y=2x+1$ is a slant asymptote in both directions. (This had to be a line: the degrees differ by exactly one.)

(b) The denominator vanishes only at $x=2$, and there the numerator is $8-6+1=3\ne 0$, so nothing cancels and $x=2$ is a vertical asymptote. From the divided form, the sign is carried entirely by $\frac{3}{x-2}$:

$$\underset{x\to 2^{+}}{\lim }f(x)=+\infty ,\underset{x\to 2^{-}}{\lim }f(x)=-\infty .$$

Intercepts: $2x^2-3x+1=(2x-1)(x-1)$, so $f(x)=0$ at $x=\frac{1}{2}$ and $x=1$ (neither equals $2$, so both are genuine). The $y$-intercept is $f(0)=\frac{1}{-2}=-\frac{1}{2}$.

(c) The sign of the remainder decides it. For $x>2$, $\frac{3}{x-2}>0$, so the curve lies above $y=2x+1$; for $x<2$, $\frac{3}{x-2}<0$, so it lies below.

The sketch has two branches. On $x<2$: far to the left the curve runs just below the line (at $x=-2$, $f=-3.75$ against the line's $-3$), rises through the roots $x=\frac{1}{2}$ and $x=1$ — with a small local maximum between them, about $0.1$ near $x=0.75$ — passes $(0,-\frac{1}{2})$ on the way, and then dives to $-\infty$ at the wall $x=2$. On $x>2$: the curve descends from $+\infty$ and settles onto the line from above (at $x=3$, $f=10$ against the line's $7$; at $x=6$, $f=13.75$ against $13$).

Draw $x=2$ and $y=2x+1$ dashed, and mark $(\frac{1}{2},0)$, $(1,0)$ and $(0,-\frac{1}{2})$.

(a) The difference law is being applied. Its hypothesis is that the two individual limits exist as real numbers; neither does, so the law is silent. The symbol $\infty -\infty$ is not a subtraction of numbers at all — it is a label for the case in which the laws refuse to answer, and "indeterminate" means "undecided", never "does not exist".

Correct treatment: for every $x\ne 0$ the expression $\frac{1}{x}-\frac{1}{x}$ is identically $0$, and a limit never inspects $x=0$ itself. So the function agrees with the constant $0$ on a punctured neighbourhood, and ${\lim }_{x\to 0}\left(\frac{1}{x}-\frac{1}{x}\right)=0$. The stated conclusion is false.

(b) The product law again requires both limits to exist as real numbers, and $+\infty$ is not one. The form $0\cdot \infty$ is indeterminate for the same reason as in (a): it records that one factor shrinks and the other grows, without recording how fast, and the rate is the whole question.

Correct treatment: $\frac{\sin x}{x}$ is the fundamental trigonometric limit. From the unit-circle area comparison one gets $\cos x\le \frac{\sin x}{x}\le 1$ for $0<x<\frac{\pi }{2}$, and the Squeeze Theorem gives ${\lim }_{x\to 0^{+}}\frac{\sin x}{x}=1$. The stated value $0$ is false.

(c) Two errors, one after the other.

First, $f$ and $g$ are not the same function. They agree for $x\ne 3$, but $f$ is undefined at $3$ while $g(3)=6$, and functions with different domains are different functions. What the theorem "limits ignore the point itself" transfers from $g$ to $f$ is the limit, never the value; this is exactly why the limit sign must stay attached while you cancel.

Second, continuity at $3$ requires all three of: $f(3)$ defined, the limit existing, and the two agreeing. The very first condition fails, so $f$ is not continuous at $3$.

Correct treatment: ${\lim }_{x\to 3}f(x)={\lim }_{x\to 3}(x+3)=6$, and $f$ has a removable discontinuity at $3$ — removable precisely because defining $f(3)=6$ repairs it. This is the one of the three whose final sentence is closest to something true, and it is still wrong as stated.

(a) The base tends to $1$ and the exponent to $+\infty$: the form is $1^{\infty }$, which is indeterminate — a base slightly above or below $1$ raised to a huge power can do anything. Take logarithms. For $0<|x|<\pi$ we have $\frac{\sin x}{x}>0$, so with

$$y={\left(\frac{\sin x}{x}\right)}^{1/x^2},\ln y=\frac{\ln \left(\frac{\sin x}{x}\right)}{x^2},$$

the logarithm is defined near $0$ and the quotient is of the form $\frac{0}{0}$, since $\frac{\sin x}{x}\to 1$ and $\ln 1=0$.

Apply the rule. Writing the numerator as $\ln (\sin x)-\ln x$, it differentiates to $\cot x-\frac{1}{x}$, and the denominator differentiates to $2x$ (nonzero for $x\ne 0$). So

$$\underset{x\to 0}{\lim }\ln y=\underset{x\to 0}{\lim }\frac{\cot x-\frac{1}{x}}{2x}=\underset{x\to 0}{\lim }\frac{x\cos x-\sin x}{2x^2\sin x}.$$

Peel off a factor whose limit is already known: $2x^2\sin x=2x^3\cdot \frac{\sin x}{x}$ and $\frac{\sin x}{x}\to 1$, so it is enough to evaluate

$$\underset{x\to 0}{\lim }\frac{x\cos x-\sin x}{2x^3},$$

still $\frac{0}{0}$. One more application: the numerator differentiates to $\cos x-x\sin x-\cos x=-x\sin x$ and the denominator to $6x^2$, giving

$$\underset{x\to 0}{\lim }\frac{-x\sin x}{6x^2}=-\frac{1}{6}\underset{x\to 0}{\lim }\frac{\sin x}{x}=-\frac{1}{6}.$$

Hence $\ln y\to -\frac{1}{6}$, and since $\exp$ is continuous, $y=e^{\ln y}\to e^{-1/6}$. The limit is

$$\boxed{e^{-1/6}}\approx \mathrm{0.84648.}$$

Stopping at $-\frac{1}{6}$ would answer a different question — that is the classic slip in a $1^{\infty }$ problem.

Sanity check: at $x=0.005$ the original expression is $0.8464816$, against $e^{-1/6}=0.8464817$.

(b) Base case $k=1$. As $x\to \infty$ both $\ln x\to \infty$ and $x\to \infty$, so $\frac{\ln x}{x}$ is of the form $\frac{\infty }{\infty }$; both functions are differentiable on $(1,\infty )$ and the denominator's derivative is $1\ne 0$. L'Hôpital's Rule gives

$$\underset{x\to \infty }{\lim }\frac{\ln x}{x}=\underset{x\to \infty }{\lim }\frac{1/x}{1}=\underset{x\to \infty }{\lim }\frac{1}{x}=0.$$

Inductive step. Let $k\ge 2$ and suppose ${\lim }_{x\to \infty }\frac{(\ln x{)}^{k-1}}{x}=0$. Since $\ln x\to \infty$, the quotient $\frac{(\ln x{)}^k}{x}$ is again of the form $\frac{\infty }{\infty }$, and the same differentiability and $g^{\prime }\ne 0$ checks apply on $(1,\infty )$. The rule gives

$$\underset{x\to \infty }{\lim }\frac{(\ln x{)}^k}{x}=\underset{x\to \infty }{\lim }\frac{k(\ln x{)}^{k-1}\cdot \frac{1}{x}}{1}=k\underset{x\to \infty }{\lim }\frac{(\ln x{)}^{k-1}}{x}=k\cdot 0=0,$$

by the inductive hypothesis — and the rule is legitimately applied because that limit on the right does exist. By induction the statement holds for every positive integer $k$. $■$

This is the hierarchy $\ln x\ll x^r\ll e^x$ made precise in the case $r=1$: no power of the logarithm can keep up with $x$ itself.