L'Hôpital's Rule and Indeterminate Forms

Main idea

When a fraction turns into $\frac00$ or $\frac\infty\infty$ , what matters is how fast the top and bottom are changing, so you can compare their derivatives instead. Every other indeterminate form can be rewritten into one of those two.

\lim\frac{f(x)}{g(x)} = \lim{\htmlData{sym=0}{\frac{f'(x)}{g'(x)}}} \quad\text{when the first limit is } \tfrac00 \text{ or } \tfrac{\pm\infty}{\pm\infty}

What the symbols mean

$\frac{f'(x)}{g'(x)}$
Quotient of derivatives
Differentiate top and bottom separately. This is not the quotient rule.

Explanation

L'Hôpital's rule is simply a shortcut for limits where the top and bottom of a fraction both head to $0$ (or both blow up), so you can't tell what the fraction does. ("Indeterminate" means exactly that: the form alone doesn't determine the answer.) All we're really saying is: if both are shrinking to zero, the one shrinking faster wins, and "how fast" is what a derivative measures. The formula reads as: $\lim\frac{f}{g}$ (the limit you're stuck on) equals $\lim\frac{f'}{g'}$ (the same limit with the top and bottom each differentiated on their own). That's all there is to it.

Think about why it works. Zoom in close enough on a smooth curve and it looks like its tangent line (we'll lean on that idea again in the next lesson). If $f(a) = g(a) = 0$ , then near $a$ , $f(x) \approx f'(a)(x - a)$ and $g(x) \approx g'(a)(x - a)$ . Divide them and the $(x - a)$ cancels, leaving $\frac{f'(a)}{g'(a)}$ . So the ratio of the functions turns into the ratio of their slopes.

Check the form first. L'Hôpital only applies to $\frac00$ and $\frac{\infty}{\infty}$ . Applying it elsewhere gives wrong answers.

A common mistake. It's tempting to use the quotient rule here, because there's a quotient and you're taking derivatives. It isn't the quotient rule: differentiate the top and the bottom separately and divide, $\frac{f'}{g'}$ , not $\left(\frac fg\right)'$ .

Repeat if needed. If the new limit is still indeterminate, apply it again.

Other indeterminate forms:

$0\cdot\infty$ : rewrite $fg$ as $\dfrac{f}{1/g}$ .
$\infty - \infty$ : combine into one fraction.
$1^\infty$ , $0^0$ , $\infty^0$ : take the log. If $y = f^g$ , then $\ln y = g\ln f$ ; find that limit $L$ , and the answer is $e^L$ .

This is how we prove which functions grow faster: $\frac{x^2}{e^x} \to \frac{2x}{e^x} \to \frac{2}{e^x} \to 0$ , so $e^x$ eventually leaves $x^2$ far behind.

See it

These are f(x) = sin 3x and g(x) = 2x, both shrinking to 0 as x closes in on 0, so f/g looks like 0/0. Notice that near 0, sin 3x is almost the straight line y = 3x. So the ratio behaves like 3x/2x, and as you can see, it settles toward 3/2 = 1.5, the ratio of the two slopes. That's L'Hôpital's rule in one picture.

Worked examples

Example 1

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

1
As $x \to 0^+$ , $x \to 0$ and $\ln x \to -\infty$ , so this is $0\cdot(-\infty)$ . Rewrite it as a fraction, $\dfrac{\ln x}{1/x}$ , which has the form $\frac{-\infty}{\infty}$ .

Example 2

Find $\displaystyle\lim_{x\to\infty}\left(1 + \frac1x\right)^x$ .

1
The base goes to $1$ and the exponent to $\infty$ , so it's $1^\infty$ . Take the log: with $y = \left(1 + \frac1x\right)^x$ , $\ln y = \dfrac{\ln(1 + 1/x)}{1/x}$ , which has the form $\frac00$ .

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

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

Problem 2+5 XP

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

Problem 3+5 XP

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

Problem 4+5 XP

Find $\displaystyle\lim_{x\to\infty}\left(1 + \frac3x\right)^x$ .

Problem 5+5 XP

Find $\displaystyle\lim_{x\to1}\frac{\ln x}{x - 1}$ .

Problem 6+5 XP

In a drug model with absorption rate $k$ and elimination rate $1$ , the concentration is $C(t) = \frac{e^{-t} - e^{-kt}}{k - 1}$ , a formula that breaks down when $k = 1$ . Find $\lim_{k\to1}C(2)$ , the concentration at $t = 2$ in that case.

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

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

Where it's used

L'Hôpital's rule settles which algorithm wins at scale ( $n\log n$ versus $n^{1.1}$ ), and gives continuous compounding: $\left(1 + \frac rn\right)^{n} \to e^r$ .

How Algorithms ScaleAlgebra · Computer Science

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