L'Hôpital's Rule¶

Theorem¶

Let $x$ approach $a\in\mathbb{R}\cup\{-\infty,+\infty\}$ through an interval. Assume that, sufficiently near $a$:

• $f$ and $g$ are differentiable, and $g'(x)\neq0\text{;}$
• either $f(x)\to0$ and $g(x)\to0\text{,}$ or $|f(x)|\to\infty$ and $|g(x)|\to\infty\text{;}$
• $\displaystyle\lim_{x\to a}\frac{f'(x)}{g'(x)}=L\text{,}$ where $L$ may be finite or infinite.

Then

Interpretation¶

L’Hôpital’s rule evaluates limits of quotients that have the indeterminate form $0/0$ or $\infty/\infty$ by comparing the rates of change of the numerator and denominator. It replaces the limit of $f/g$ by the limit of $f'/g'$, which is often simpler.

$\displaystyle\lim_{x\to0}\frac{f(x)}{g(x)}=\lim_{x\to0}\frac{f'(x)}{g'(x)}=\frac{f'(0)}{g'(0)}=\frac{1}{2}$$\frac{1}{2}$: near $0$ both functions are close to their tangents (dashed), with slopes $f'(0)=1$ and $g'(0)=2$$2$, so the quotient tends to the ratio of the slopes.

Each hypothesis matters. The rule needs an indeterminate form: $\frac{x+1}{x+2}\to\frac{1}{2}$ as $x\to0$$0$, while the quotient of the derivatives is $1$. It also needs the limit of $f'/g'$ to exist: $\frac{x+\sin x}{x}\to1$ as $x\to\infty$$\infty$, but $\frac{f'(x)}{g'(x)}=1+\cos x$ has no limit. In such a case the rule gives no information, and the limit of $f/g$ must be found in another way.

The rule may be applied again as long as the form stays indeterminate: as $x\to\infty$$\infty$, the quotient $x^2/\mathrm{e}^x$ leads to $2x/\mathrm{e}^x$ and then to $2/\mathrm{e}^x\to0$$0$. Other indeterminate forms are first rewritten as a quotient of type $0/0$ or $\infty/\infty$.

A common slip is to differentiate the quotient $f/g$ with the quotient rule. The rule differentiates numerator and denominator separately: it uses $f'/g'$, not $(f/g)'$.

Key Results¶

• L’Hôpital’s rule applies only to indeterminate forms of type $0/0$ or $\infty/\infty\text{.}$
• Differentiability of both numerator and denominator is required in a punctured neighbourhood of the point.
• The rule must not be used to justify limits that are needed to establish the derivatives involved.
• The rule needs $\frac{f'(x)}{g'(x)}$ to have a limit $L\text{,}$ finite or an infinite extended real number; if that limit does not exist, the rule gives no conclusion about $\frac{f(x)}{g(x)}\text{.}$

Examples¶

• The limit $\displaystyle \lim_{x\to\infty}\frac{\ln x}{x}=0$ can be evaluated using L’Hôpital’s rule.

• Although formally applicable, L’Hôpital’s rule cannot be used to justify

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

  because the derivative of $\sin x$ depends on this limit itself. This would result in circular reasoning.

Prerequisites¶

• Indeterminate Form
• Derivative and Differentiability
• Extended Real Number

Related Concepts¶

• Limit at a Point
• Taylor's Theorem

Sources¶

• Math I: Calculus · Lecture 8

Aliases: L'Hospital's rule, L'Hopital's rule, Bernoulli's rule Symbols: element of, infinity, real numbers, union

Last updated: 27 October 2026 11:52