L'Hôpital's Rule and Indeterminate Forms

(a) The denominator tends to $1\neq0$, so substitute: $\frac{0}{1}=0$. L'Hôpital would give $\frac{\cos x}{1}\to1$, which is simply wrong.

(b) Form $\frac{\infty}{\infty}$: the rule gives $\lim\frac{1/x}{1}=0$ (or quote the growth hierarchy).

(c) Numerator $\to1$, denominator $\to0^{+}$: determinate and infinite, so the limit is $+\infty$.

(d) Form $1^{\infty}$. With $y=\left(1+\frac{3}{x}\right)^{x}$ and $t=\frac{1}{x}$:

$\ln y=\frac{\ln(1+3t)}{t}\ \left(\tfrac{0}{0}\right),\qquad \lim_{t\to0^{+}}\frac{3/(1+3t)}{1}=3$

so the limit is $e^{3}$.

Form: $e^{2x}-1\to0$ and $\sin 3x\to0$, so $\frac{0}{0}$. Both functions are differentiable everywhere and $g'(x)=3\cos 3x\neq0$ on $\left(-\frac{\pi}{6},\frac{\pi}{6}\right)$, so the rule applies there.

$\lim_{x\to0}\frac{e^{2x}-1}{\sin 3x}=\lim_{x\to0}\frac{2e^{2x}}{3\cos 3x}=\frac{2}{3}$

The new quotient is continuous at $0$ with denominator $3\neq0$, so nothing is indeterminate and we stop.

Check without the rule: $\frac{e^{2x}-1}{2x}\to1$ and $\frac{\sin 3x}{3x}\to1$, so the quotient behaves like $\frac{2x}{3x}$.

Form $\frac{0}{0}$, and $g'(x)=3x^{2}\neq0$ for $x\neq0$.

$\lim_{x\to0}\frac{x-\sin x}{x^{3}}=\lim_{x\to0}\frac{1-\cos x}{3x^{2}}=\lim_{x\to0}\frac{\sin x}{6x}=\frac{1}{6}$

Each equality needs its own check: after stage 1 the form is still $\frac{0}{0}$, and so is $\frac{\sin x}{6x}$ — but $\lim_{x\to0}\frac{\sin x}{x}=1$ is standard, so stop there. A third application differentiates $\sin x$, which is circular if the standard limit is what you were asked to establish.

Check: $\sin x=x-\frac{x^{3}}{6}+\cdots$, so the quotient is $\frac{1}{6}-\frac{x^{2}}{120}+\cdots$.

Both pieces blow up, so the form is $\infty-\infty$. Combine (valid for $x>0$, $x\neq1$):

$\frac{x}{x-1}-\frac{1}{\ln x}=\frac{x\ln x-x+1}{(x-1)\ln x}$

Top and bottom both tend to $0$ at $x=1$: form $\frac{0}{0}$. Differentiating separately, $N'(x)=\ln x$ and $D'(x)=\ln x+1-\frac{1}{x}$, and both still tend to $0$, so apply the rule once more:

$\lim_{x\to1}\frac{1/x}{\frac{1}{x}+\frac{1}{x^{2}}}=\frac{1}{1+1}=\frac{1}{2}$

That last quotient is determinate, so stop.

Base $\to1$, exponent $\to+\infty$: form $1^{\infty}$. For $x>1$ the base is positive, so put $y=\left(\frac{x+2}{x-1}\right)^{3x}$ and move $3x$ downstairs:

$\ln y=\frac{\ln(x+2)-\ln(x-1)}{1/(3x)}\qquad\left(\tfrac{0}{0}\right)$

The top differentiates to $\frac{1}{x+2}-\frac{1}{x-1}=\frac{-3}{(x+2)(x-1)}$ and the bottom to $-\frac{1}{3x^{2}}$, so

$\lim_{x\to\infty}\ln y=\lim_{x\to\infty}\frac{9x^{2}}{(x+2)(x-1)}=9$

Since $\exp$ is continuous, the limit is $e^{9}$ — not $9$.

The form is $\frac{\infty}{\infty}$, so the rule is legal. One application gives $\frac{4x}{3\sqrt{4x^{2}+1}}$; applying it to that gives $\frac{4}{12x/\sqrt{4x^{2}+1}}=\frac{\sqrt{4x^{2}+1}}{3x}$ — where you began. Every step is valid, but the process never ends.

Use algebra instead: for $x>0$, $\sqrt{4x^{2}+1}=x\sqrt{4+\frac{1}{x^{2}}}$, so

$\frac{\sqrt{4x^{2}+1}}{3x+5}=\frac{\sqrt{4+\frac{1}{x^{2}}}}{3+\frac{5}{x}}\to\frac{\sqrt{4}}{3}=\frac{2}{3}$

As $x\to-\infty$ the same algebra gives $-\frac{2}{3}$, because there $\sqrt{x^{2}}=-x$.