# L'Hôpital's Rule: Formula, Conditions and Examples

> L'Hôpital's rule evaluates 0/0 and ∞/∞ limits by differentiating the top and bottom. The exact conditions, worked examples and other indeterminate forms.

Source: https://practicecalculus.com/rules/lhopitals-rule
Updated: 2026-11-03

$$ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} $$

## At a glance

- **Applies to:** Limits of the indeterminate form $\frac{0}{0}$ or $\frac{\pm\infty}{\pm\infty}$
- **In words:** Differentiate the numerator and the denominator separately, then take the limit again
- **Also requires:** The new limit $\lim \frac{f'(x)}{g'(x)}$ exists or is $\pm\infty$
- **Works for:** $x \to a$, one-sided limits and $x \to \pm\infty$
- **Other forms:** $0 \cdot \infty$, $\infty - \infty$, $1^\infty$, $0^0$, $\infty^0$ after rewriting

## Key takeaways

- Use L'Hôpital's rule only on $\frac{0}{0}$ or $\frac{\infty}{\infty}$. Check the form first, every time.
- Differentiate the top and the bottom separately. This is not the quotient rule.
- You may apply the rule repeatedly while the form stays indeterminate.
- Rewrite $0 \cdot \infty$, $\infty - \infty$ and exponent forms as a fraction before using it.

**L'Hôpital's rule** evaluates limits of fractions that come out as $\frac{0}{0}$ or $\frac{\infty}{\infty}$. Instead of the original fraction, you take the limit of the derivative of the top over the derivative of the bottom. It is often the fastest way through a limit that algebra cannot easily simplify.

## The rule and its conditions

Suppose $f$ and $g$ are differentiable on an open interval around $a$ (except possibly at $a$), and $g'(x) \ne 0$ there for $x \ne a$. If

$$\lim_{x \to a} f(x) = \lim_{x \to a} g(x) = 0 \quad \text{or} \quad \lim_{x \to a} f(x) = \pm\infty,\ \lim_{x \to a} g(x) = \pm\infty$$

then

$$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$$

provided the limit on the right exists or is $\pm\infty$. The same holds for one-sided limits and for $x \to \infty$ or $x \to -\infty$.

If the right-hand limit does not exist, the rule tells you nothing. The original limit may still exist.

```graph
title: 'Graph of y = sin(x)/x approaching 1 at x = 0'
caption: 'At $x = 0$ the fraction is $\frac{0}{0}$. L''Hôpital''s rule gives $\lim_{x \to 0} \frac{\cos x}{1} = 1$, matching the graph.'
x: [-10, 10]
y: [-0.5, 1.3]
functions:
  - expr: sin(x)/x
    label: y = sin(x)/x
points:
  - [0, 1, '(0, 1)', open]
```

## Worked examples

**Example 1: $\frac{0}{0}$.** At $x = 0$, $\frac{\sin x}{x}$ is $\frac{0}{0}$:

$$\lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{\cos x}{1} = 1$$

(The derivative of $\sin x$ is itself proved using this limit, so in a proof course you would use the squeeze theorem instead. For computation, the shortcut is fine.)

**Example 2: applying the rule twice.** Each step is still $\frac{0}{0}$ until the last:

$$\lim_{x \to 0} \frac{e^x - 1 - x}{x^2} = \lim_{x \to 0} \frac{e^x - 1}{2x} = \lim_{x \to 0} \frac{e^x}{2} = \frac{1}{2}$$

**Example 3: $\frac{\infty}{\infty}$.** As $x \to \infty$, both $\ln x$ and $x$ grow without bound:

$$\lim_{x \to \infty} \frac{\ln x}{x} = \lim_{x \to \infty} \frac{1/x}{1} = 0$$

So $x$ grows faster than $\ln x$.

**Example 4: $0 \cdot \infty$.** Rewrite the product as a fraction first. As $x \to 0^+$, $x \ln x$ has the form $0 \cdot (-\infty)$:

$$\lim_{x \to 0^+} x \ln x = \lim_{x \to 0^+} \frac{\ln x}{1/x} = \lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} (-x) = 0$$

## Other indeterminate forms

| Form | How to rewrite |
| --- | --- |
| $0 \cdot \infty$ | Write $fg$ as $\frac{f}{1/g}$ or $\frac{g}{1/f}$ |
| $\infty - \infty$ | Combine into one fraction (common denominator or conjugate) |
| $1^\infty$, $0^0$, $\infty^0$ | Let $y = f^g$, take $\ln y = g \ln f$, find that limit, then exponentiate |

For example, with $y = \left(1 + \frac{1}{x}\right)^x$ you get $\ln y = \frac{\ln(1 + 1/x)}{1/x}$, which is $\frac{0}{0}$ as $x \to \infty$. The rule gives $\ln y \to 1$, so $y \to e$.

## Common mistakes

1. **Using the rule when the form is not indeterminate.** $\lim_{x \to 0} \frac{x + 1}{x + 2} = \frac{1}{2}$ by [direct substitution](/rules/direct-substitution). Differentiating would wrongly give $1$.
2. **Using the quotient rule.** Differentiate $f$ and $g$ separately. Do not compute the derivative of $\frac{f}{g}$ with the [quotient rule](/rules/quotient-rule).
3. **Skipping the rewrite.** The rule only applies to a fraction. Turn products, differences and powers into a fraction first.
4. **Not rechecking the form.** Before each repeated application, confirm the new limit is still $\frac{0}{0}$ or $\frac{\infty}{\infty}$.

For more examples, including limits at infinity and exponent forms, read the full guide to [L'Hôpital's rule](/blog/lhopitals-rule).

## Frequently asked questions

### When can you use L'Hôpital's rule?

When direct substitution gives $\frac{0}{0}$ or $\frac{\pm\infty}{\pm\infty}$, $f$ and $g$ are differentiable near the point, $g'(x) \ne 0$ near it, and the limit of $\frac{f'}{g'}$ exists or is infinite.

### Is L'Hôpital's rule the same as the quotient rule?

No. L'Hôpital's rule takes the derivative of the numerator and the derivative of the denominator separately. The quotient rule finds the derivative of the whole fraction and has nothing to do with this limit.

### Can you use L'Hôpital's rule more than once?

Yes. If the new limit is still $\frac{0}{0}$ or $\frac{\infty}{\infty}$, apply the rule again. Check the form before every application.

### How do you use L'Hôpital's rule for 1 to the infinity?

Take the natural log to turn the power into a product, rewrite that product as a fraction, apply the rule, and then exponentiate the result.