Practice: L'Hôpital's Rule

(a) $\underset{x\to 0}{lim}{\displaystyle \frac{e^{2x}-1}{\sin x}}$. Form. At $x=0$: numerator $e^0-1=0$, denominator $\sin 0=0$. The form is $\frac{0}{0}$. Both functions are differentiable near $0$ and $\frac{d}{dx}\sin x=\cos x$ is nonzero on a punctured neighbourhood of $0$, so the hypotheses hold and the rule applies. An indeterminate form on its own does not license it. Apply. Differentiate separately, $\frac{d}{dx}(e^{2x}-1)=2e^{2x}$ by the chain rule, and $\frac{d}{dx}\sin x=\cos x$:

Check. The new quotient is determinate at 0, so one application suffices. Numerically the original reads $2.1034$ at $x=0.1$ and $2.0100$ at $x=0.01$, approaching 2 ✓. The answer is the ratio of rates: the numerator leaves zero at speed 2, the denominator at speed 1. (b) $\underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{x^3}{e^x}}$. Form. Both parts grow without bound: $\frac{\mathrm{\infty }}{\mathrm{\infty }}$ ✓. | Pass | Expression | Form | Continue? |
|---|---|---|---|
| — | ${\displaystyle \frac{x^3}{e^x}}$ | $\mathrm{\infty }/\mathrm{\infty }$ | yes |
| 1 | ${\displaystyle \frac{3x^2}{e^x}}$ | $\mathrm{\infty }/\mathrm{\infty }$ | yes |
| 2 | ${\displaystyle \frac{6x}{e^x}}$ | $\mathrm{\infty }/\mathrm{\infty }$ | yes |
| 3 | ${\displaystyle \frac{6}{e^x}}$ | constant over $\mathrm{\infty }$ | no | Three applications. After the third the numerator is the constant 6 while the denominator grows without bound, so the form is no longer indeterminate and the limit is read directly:

Where to stop. A fourth application would differentiate the constant to 0, giving $\frac{0}{e^x}$. The same answer by luck here, but the habit is the error of applying the rule to a determinate form, which changes the answer in other problems. In general $\frac{x^n}{e^x}$ needs exactly $n$ applications, which is why any exponential beats any power. (c) $\underset{x\to 0^{+}}{lim}{x}^2\ln x$. Form. $x^2\to 0$ and $\ln x\to -\mathrm{\infty }$, so the form is $0\cdot (-\mathrm{\infty })$. A product, which the rule does not reach directly. Rewrite. Move the factor with the easier reciprocal into the denominator:

now of the form $\frac{-\mathrm{\infty }}{\mathrm{\infty }}$ ✓. Apply. $\frac{d}{dx}\ln x=\frac{1}{x}$ and $\frac{d}{dx}{x}^{-2}=-2x^{-3}$, so

Check. $x^2\ln x$ reads $-0.0230$ at $x=0.1$ and $-0.00046$ at $x=0.01$, approaching 0 ✓. The other rewrite would be $\frac{x^2}{1/\ln x}$, whose denominator differentiates to $-\frac{1}{x{\ln }^{2}x}$, valid, and worse than what we started with. (d) $\underset{x\to 1}{lim}{\displaystyle \frac{x^2+3}{x+1}}$. Form. At $x=1$: numerator $1+3=4$, denominator $2$. The form is $\frac{4}{2}$, not indeterminate. Therefore the rule does not apply. The quotient is continuous at 1, so substitution answers it:

What misapplying the rule would give. Differentiating top and bottom gives $\frac{2x}{1}$, which at $x=1$ is 2. The same number, by coincidence. That coincidence should not be relied on: for $\frac{x+1}{x+2}$ at $x=0$ the correct answer is $\frac{1}{2}$ and the rule returns 1. The discipline is to check the form, not to check whether the wrong method happened to agree. (e) $\underset{x\to \mathrm{\infty }}{lim}{\displaystyle \frac{x+\sin x}{x}}$. Form. The numerator is unbounded (since $\sin x$ is bounded and $x$ is not) and so is the denominator: $\frac{\mathrm{\infty }}{\mathrm{\infty }}$ ✓. The hypotheses hold, so the rule may legitimately be applied. Apply. Differentiating separately gives

which does not exist. The expression oscillates between 0 and 2 forever, taking both values infinitely often. What the theorem says. Its third hypothesis is that $lim{f}^{\prime }/g^{\prime }$ exists or is infinite. That hypothesis fails, so the theorem is inapplicable and asserts nothing about the original limit. What the theorem does not say. It does not say the original limit fails to exist. The implication runs one way only: if $f^{\prime }/g^{\prime }$ has a limit, then $f/g$ has the same one. The converse is false, and this is the standard counterexample. Finding the limit another way. Split the quotient:

Since $-1\le \sin x\le 1$, dividing by gives $-\frac{1}{x}\le \frac{\sin x}{x}\le \frac{1}{x}$, and both bounds tend to 0. By the squeeze theorem $\frac{\sin x}{x}\to 0$, so

Check. At $x=100$ the expression is $0.99494$, at $x=1000$ it is $1.00083$, at $x={10}^6$ it is $0.99999965$, hovering ever closer to 1 from either side while never settling exactly, which is the bounded oscillation being divided away ✓. A rule with a hypothesis about its own conclusion can be silent without being violated. Recognising silence, as distinct from a verdict, is what keeps the rule from being used to prove something false.