Practice / Calculus

L'Hôpital's rule and indeterminate forms

Ten limits by L'Hôpital's rule: 0/0 and ∞/∞, a rule applied twice, x ln x and the 0 log 0 = 0 of entropy, xˣ and (1 + 2/x)ˣ by taking logs, a difference of fractions, and a limit the rule cannot find, with worked solutions and the mistakes of using it on a limit that is not indeterminate and of reading 1^∞ as 1.

Before you start

When substitution gives $\tfrac00$ or $\tfrac\infty\infty$ L'Hôpital's rule replaces the limit of a ratio with the limit of the ratio of the derivatives. It is how the convention $0\log 0 = 0$ in entropy is justified, and why $x^2e^{-x}$ and $\tfrac{\ln x}{\sqrt x}$ go to 0: exponentials beat powers, and powers beat logs. The work goes wrong in a few places: the rule used on a limit that is not indeterminate, the quotient rule used instead of differentiating top and bottom separately, $1^\infty$ read as 1, a log taken and never undone, and a conclusion drawn when the rule does not apply. These ten problems cover each of them.

L'Hôpital's rule. Suppose $f$ and $g$ are differentiable near $a$ (except possibly at $a$ $g'(x) \neq 0$ near $a$ and $\tfrac{f(x)}{g(x)}$ has the form $\tfrac00$ or $\tfrac{\pm\infty}{\pm\infty}$ as $x \to a$ If $\lim\tfrac{f'(x)}{g'(x)}$ exists (or is $\pm\infty$ then $\lim\tfrac{f(x)}{g(x)}$ equals it. The same holds for one-sided limits and for $a = \pm\infty$
$f'$ and $g'$ are differentiated separately. It is not the quotient rule.
Check the form before every use. If substitution gives a number, that number is the limit and the rule does not apply.
The other indeterminate forms are rewritten first. $0\cdot\infty$ write the product as a quotient, $fg = \tfrac{f}{1/g}$ $\infty - \infty$ combine into one fraction. $1^\infty$ $0^0$ and $\infty^0$ take the log, find the limit $L$ of the log, and the answer is $e^L$
If $\lim\tfrac{f'}{g'}$ does not exist, the rule says nothing: the original limit may still exist (Problem 10).
Derivatives used: $(\sin x)' = \cos x$ $(\cos x)' = -\sin x$ $(e^x)' = e^x$ $(\ln x)' = \tfrac1x$ $(x^n)' = nx^{n-1}$

Builds on: Evaluating limits: factoring, conjugates and limits at infinity, Differentiation rules: chain, product, quotient

Problems

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Find $\displaystyle\lim_{x\to0}\frac{\sin 3x}{x}$
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Find $\displaystyle\lim_{x\to0}\frac{e^x - 1 - x}{x^2}$
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Find $\displaystyle\lim_{x\to\infty}\frac{x^2}{e^x}$
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Find $\displaystyle\lim_{x\to0^+} x\ln x$
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Find $\displaystyle\lim_{x\to\infty}\frac{\ln x}{\sqrt x}$
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Find $\displaystyle\lim_{x\to0}\frac{1 - \cos x}{x^2}$
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Find $\displaystyle\lim_{x\to0^+} x^x$
···
Find $\displaystyle\lim_{x\to\infty}\Big(1 + \frac2x\Big)^x$
···
Find $\displaystyle\lim_{x\to0}\Big(\frac1x - \frac1{\sin x}\Big)$
···
Find $\displaystyle\lim_{x\to\infty}\frac{x + \sin x}{x}$ What happens if you use L'Hôpital's rule?

Answers

$3$
$\tfrac12$
$0$
$0$
$0$
$\tfrac12$
$1$
$e^2 \approx 7.389$
$0$
$1$ L'Hôpital's rule does not apply, because $\lim(1 + \cos x)$ does not exist

Worked solutions

Problem 1

Find $\displaystyle\lim_{x\to0}\frac{\sin 3x}{x}$

At $x = 0$ $\tfrac{\sin 0}{0} = \tfrac00$ Indeterminate, so the rule may apply.
$\displaystyle\lim_{x\to0}\frac{3\cos 3x}{1}$ Differentiate top and bottom separately; the chain rule gives the 3.
$3$ $3\cos 0 = 3$ Near 0, $\sin 3x \approx 3x$

Problem 2

Find $\displaystyle\lim_{x\to0}\frac{e^x - 1 - x}{x^2}$

At $x = 0$ $\tfrac{1 - 1 - 0}{0} = \tfrac00$ Indeterminate.
$\displaystyle\lim_{x\to0}\frac{e^x - 1}{2x}$ again $\tfrac00$ One application. Check the form again before the next one.
$\displaystyle\lim_{x\to0}\frac{e^x}{2}$ A second application.
$\tfrac12$ $\tfrac{e^0}2$ It is the $\tfrac{x^2}{2}$ term of the Taylor series $e^x = 1 + x + \tfrac{x^2}2 + \cdots$

Problem 3

Find $\displaystyle\lim_{x\to\infty}\frac{x^2}{e^x}$

As $x \to \infty$ $\tfrac\infty\infty$ Indeterminate.
$\displaystyle\lim_{x\to\infty}\frac{2x}{e^x}$ still $\tfrac\infty\infty$ The power drops by one; the exponential does not change.
$\displaystyle\lim_{x\to\infty}\frac{2}{e^x}$ A second application. The top is now a constant.
$0$ $e^x \to \infty$ The same argument, applied $n$ times, shows $x^n e^{-x} \to 0$ for every $n$

Problem 4

Find $\displaystyle\lim_{x\to0^+} x\ln x$

As $x \to 0^+$ $0\cdot(-\infty)$ Indeterminate, but not a quotient yet.
$x\ln x = \dfrac{\ln x}{1/x}$ which has the form $\tfrac{-\infty}{\infty}$ Move $x$ to the denominator as $\tfrac1x$ Keeping $\ln x$ on top makes the derivatives simple.
$\displaystyle\lim_{x\to0^+}\frac{1/x}{-1/x^2} = \lim_{x\to0^+}(-x)$ Differentiate top and bottom; $\tfrac{1/x}{-1/x^2} = -x$
$0$ This is why entropy uses $0\log 0 = 0$ $p\log p \to 0$ as $p \to 0^+$ so an outcome of probability 0 adds nothing.

Problem 5

Find $\displaystyle\lim_{x\to\infty}\frac{\ln x}{\sqrt x}$

As $x \to \infty$ $\tfrac\infty\infty$ Indeterminate.
$\displaystyle\lim_{x\to\infty}\frac{1/x}{1/(2\sqrt x)} = \lim_{x\to\infty}\frac{2\sqrt x}{x}$ $(\sqrt x)' = \tfrac1{2\sqrt x}$
$= \displaystyle\lim_{x\to\infty}\frac{2}{\sqrt x}$ $\tfrac{\sqrt x}{x} = \tfrac1{\sqrt x}$
$0$ Any positive power of $x$ eventually outgrows $\ln x$ even $x^{1/2}$

Problem 6

Find $\displaystyle\lim_{x\to0}\frac{1 - \cos x}{x^2}$

At $x = 0$ $\tfrac{1 - 1}{0} = \tfrac00$ Indeterminate.
$\displaystyle\lim_{x\to0}\frac{\sin x}{2x}$ still $\tfrac00$ $(1 - \cos x)' = \sin x$
$\displaystyle\lim_{x\to0}\frac{\cos x}{2}$ A second application.
$\tfrac12$ $\tfrac{\cos 0}2$ So $\cos x \approx 1 - \tfrac{x^2}2$ near 0.

Problem 7

Find $\displaystyle\lim_{x\to0^+} x^x$

As $x \to 0^+$ $0^0$ Indeterminate: a base going to 0 pulls toward 0, an exponent going to 0 pulls toward 1.
Let $y = x^x$ then $\ln y = x\ln x$ Take the log to bring the exponent down.
$\displaystyle\lim_{x\to0^+}\ln y = 0$ Problem 4.
$1$ $y = e^{\ln y}$ and $e^x$ is continuous, so $y \to e^0 = 1$ The answer is not 0, which is the limit of the log (mistake 4).

Problem 8

Find $\displaystyle\lim_{x\to\infty}\Big(1 + \frac2x\Big)^x$

As $x \to \infty$ $1^\infty$ Indeterminate: the base goes to 1, but the exponent grows without bound.
$\ln y = x\ln\Big(1 + \dfrac2x\Big) = \dfrac{\ln(1 + 2/x)}{1/x}$ which has the form $\tfrac00$ Take the log, then write the product as a quotient.
$\displaystyle\lim_{x\to\infty}\frac{\dfrac{1}{1 + 2/x}\cdot\Big(-\dfrac{2}{x^2}\Big)}{-\dfrac1{x^2}} = \lim_{x\to\infty}\frac{2}{1 + 2/x} = 2$ The chain rule on the top: $\big(\tfrac2x\big)' = -\tfrac2{x^2}$ The $-\tfrac1{x^2}$ cancels.
$e^2 \approx 7.389$ $y = e^{\ln y} \to e^2$ This is compound interest at a rate of 200% compounded ever more often.

Problem 9

Find $\displaystyle\lim_{x\to0}\Big(\frac1x - \frac1{\sin x}\Big)$

As $x \to 0^+$ $\infty - \infty$ Indeterminate; from the left it is $-\infty + \infty$
$\dfrac1x - \dfrac1{\sin x} = \dfrac{\sin x - x}{x\sin x}$ which has the form $\tfrac00$ Combine into one fraction over the common denominator.
$\displaystyle\lim_{x\to0}\frac{\cos x - 1}{\sin x + x\cos x}$ still $\tfrac00$ The product rule on the bottom: $(x\sin x)' = \sin x + x\cos x$
$\displaystyle\lim_{x\to0}\frac{-\sin x}{2\cos x - x\sin x} = \frac{0}{2}$ Again: $(\sin x + x\cos x)' = \cos x + \cos x - x\sin x$
$0$ From both sides. Near 0, $\sin x \approx x - \tfrac{x^3}6$ so $\tfrac1{\sin x} \approx \tfrac1x\big(1 + \tfrac{x^2}6\big)$ and the difference is about $-\tfrac x6$

Problem 10

Find $\displaystyle\lim_{x\to\infty}\frac{x + \sin x}{x}$ What happens if you use L'Hôpital's rule?

As $x \to \infty$ $\tfrac\infty\infty$ Indeterminate.
The rule gives $\displaystyle\lim_{x\to\infty}\frac{1 + \cos x}{1}$ which does not exist. $1 + \cos x$ keeps swinging between 0 and 2. The rule needs this limit to exist, so it gives no answer here, and it does not show that the original limit fails to exist.
$\dfrac{x + \sin x}{x} = 1 + \dfrac{\sin x}{x}$ Divide top and bottom by $x$ instead.
$-\dfrac1x \le \dfrac{\sin x}{x} \le \dfrac1x$ so $\dfrac{\sin x}{x} \to 0$ $\lvert\sin x\rvert \le 1$ and both bounds go to 0 (squeeze).
$1$ L'Hôpital's rule does not apply, because $\lim(1 + \cos x)$ does not existThe rule is a sufficient condition, not a test: when the ratio of derivatives has no limit, use another method.

Where this goes wrong

1. Using the rule on a limit that is not indeterminate

The rule is used so often that it becomes the first step for every fraction.

$\displaystyle\lim_{x\to1}\frac{x^2 + 1}{x + 1}$ A limit to find.
“It is a fraction, so differentiate top and bottom.”The habit that causes the mistake: skipping the check of the form.
$\displaystyle\lim_{x\to1}\frac{2x}{1} = 2$ At $x = 1$ the fraction is $\tfrac22 = 1$ not $\tfrac00$ or $\tfrac\infty\infty$ so the rule does not apply, and the limit is 1 by substitution. Check the form every time, including before each repeat application.

2. The quotient rule instead of two separate derivatives

L'Hôpital's rule involves a fraction and derivatives, and the quotient rule is the derivative of a fraction.

$\displaystyle\lim_{x\to0}\frac{\sin 3x}{x}$ form $\tfrac00$ Right so far: Problem 1, step 1.
“Differentiate the fraction.”The habit that causes the mistake: the rule remembered as "take the derivative".
$\displaystyle\lim_{x\to0}\frac{3x\cos 3x - \sin 3x}{x^2} = 0$ That is the limit of the derivative of $\tfrac{\sin 3x}{x}$ the slope of its graph near 0, which is a different question. The rule differentiates the top and the bottom separately: $\tfrac{3\cos 3x}{1} \to 3$

3. Reading 1^∞ as 1

One to any power is one, so a base that goes to 1 seems to settle the limit.

$\displaystyle\lim_{x\to\infty}\Big(1 + \frac2x\Big)^x$ Right so far: Problem 8.
“The base goes to 1, and $1^x = 1$ ”The habit that causes the mistake: letting the base reach its limit while the exponent is still finite.
$\displaystyle\lim_{x\to\infty}\Big(1 + \frac2x\Big)^x = 1$ The base is never exactly 1, and the exponent grows as fast as the base approaches 1. At $x = 100$ the value is already $1.02^{100} \approx 7.24$ Take the log: $x\ln\big(1 + \tfrac2x\big) \to 2$ so the limit is $e^2 \approx 7.389$

4. Forgetting to undo the log

After taking a log, the limit that comes out looks like the end of the problem.

$y = x^x$ so $\ln y = x\ln x \to 0$ as $x \to 0^+$ Right so far: Problem 7, steps 2 and 3.
“The limit is 0.”The habit that causes the mistake: reporting the limit of $\ln y$ as the limit of $y$
$\displaystyle\lim_{x\to0^+} x^x = 0$ $0$ is the limit of $\ln y$ The limit of $y$ is $e^0 = 1$ A check: $0.01^{0.01} \approx 0.955$ near 1 and nowhere near 0.

5. Concluding the limit does not exist

When the rule produces something with no limit, it looks like the answer is "no limit".

$\displaystyle\lim_{x\to\infty}\frac{x + \sin x}{x}$ form $\tfrac\infty\infty$ Right so far: Problem 10.
“L'Hôpital: $\lim(1 + \cos x)$ does not exist.”Right so far as a fact about $1 + \cos x$
So $\displaystyle\lim_{x\to\infty}\frac{x + \sin x}{x}$ does not existThe rule only says that if the ratio of derivatives has a limit, the original ratio has the same one. It says nothing when the ratio of derivatives has no limit. Here $\tfrac{x + \sin x}{x} = 1 + \tfrac{\sin x}{x} \to 1$

Print this set: lhopitals-rule.pdf (problems, answers, and worked solutions on separate pages).

Read the guide: L'Hôpital's rule: check the form, then differentiate top and bottom separately