Many limits have the form $\frac{0}{0}$ but cannot be factored, like $\frac{e^x - 1}{x}$ at 0. Derivatives, including the Derivatives of Exponential and Log Functions, give a general method. It also handles the $\frac{\infty}{\infty}$ forms from Limits at Infinity and Horizontal Asymptotes.
Key idea
L'Hôpital's rule. If $\lim f(x) = \lim g(x) = 0$, or both are $\pm\infty$, then $$\lim \frac{f(x)}{g(x)} = \lim \frac{f'(x)}{g'(x)},$$ provided the right side exists (or is $\pm\infty$). Differentiate the top and the bottom separately. The rule works for $x \to a$, one-sided limits, and $x \to \pm\infty$.
Why it works
Near $x = a$, with $f(a) = g(a) = 0$, both functions are close to their tangent lines: $f(x) \approx f'(a)(x - a)$ and $g(x) \approx g'(a)(x - a)$. So $$\frac{f(x)}{g(x)} \approx \frac{f'(a)(x - a)}{g'(a)(x - a)} = \frac{f'(a)}{g'(a)}$$. When the zeros of top and bottom collide, what matters is how fast each one approaches 0, and that is what the derivatives measure.
Worked example: Zero over zero
Find $\displaystyle\lim_{x \to 0} \frac{e^{2x} - 1}{x}$.
- At $x = 0$: $\frac{1 - 1}{0} = \frac{0}{0}$. The rule applies.
- Derivatives: top $2e^{2x}$, bottom $1$.
- $\lim_{x \to 0} 2e^{2x} = 2$.
Answer: $2$
Worked example: Infinity over infinity, twice
Find $\displaystyle\lim_{x \to \infty} \frac{x^2}{e^x}$.
- $\frac{\infty}{\infty}$: apply the rule. $\frac{2x}{e^x}$, still $\frac{\infty}{\infty}$.
- Again: $\frac{2}{e^x} \to 0$.
Answer: $0$: exponentials outgrow powers.
Worked example: Other forms
Find $\displaystyle\lim_{x \to 0^+} x\ln x$ and $\displaystyle\lim_{x \to \infty}\left(1 + \frac{1}{x}\right)^x$.
- $x\ln x$ is $0 \cdot (-\infty)$. Rewrite as $\frac{\ln x}{1/x}$, a $\frac{-\infty}{\infty}$ form. The rule gives $\frac{1/x}{-1/x^2} = -x \to 0$.
- For $y = \left(1 + \frac{1}{x}\right)^x$ (form $1^\infty$), take logs: $\ln y = \frac{\ln(1 + 1/x)}{1/x}$, a $\frac{0}{0}$ form.
- The rule: $\frac{\frac{1}{1 + 1/x}\cdot\left(-\frac{1}{x^2}\right)}{-\frac{1}{x^2}} = \frac{1}{1 + 1/x} \to 1$. So $\ln y \to 1$ and $y \to e$.
Answer: $0$ and $e$
| Form | Indeterminate? | What to do |
|---|---|---|
| $\frac{0}{0}$, $\frac{\infty}{\infty}$ | yes | apply the rule directly |
| $0 \cdot \infty$ | yes | rewrite $fg$ as $\frac{f}{1/g}$ |
| $\infty - \infty$ | yes | combine into one fraction |
| $1^{\infty}$, $0^0$, $\infty^0$ | yes | take logs, find that limit, exponentiate |
| $\frac{1}{0}$, $\frac{\infty}{0}$ | no | an infinite limit (check signs) |
| $\frac{0}{\infty}$, $0^{\infty}$ | no | the limit is 0 |
Common mistake
Two misuses. First, applying the quotient rule to $\frac{f}{g}$; the rule uses $\frac{f'}{g'}$, not $\left(\frac{f}{g}\right)'$. Second, using it when the limit is not indeterminate: $\lim_{x \to 0}\frac{x + 1}{x + 2} = \frac{1}{2}$ by substitution, but the rule would wrongly give $\frac{1}{1} = 1$. Check the form before every application.