L’Hôpital’s rule allows us to compute limits that arise in the indeterminate forms 00\frac{0}{0} and \frac{\infty}{\infty} by passing to the ratio of the derivatives.

Theorem — L'Hôpital

If limxx0f(x)g(x)\lim_{x\to x_0}\dfrac{f(x)}{g(x)} is of the form 00\dfrac{0}{0} or \dfrac{\infty}{\infty}, and if limxx0f(x)g(x)\lim_{x\to x_0}\dfrac{f'(x)}{g'(x)} exists (finite or infinite), then: limxx0f(x)g(x)=limxx0f(x)g(x).\lim_{x\to x_0}\frac{f(x)}{g(x)} = \lim_{x\to x_0}\frac{f'(x)}{g'(x)}.

Caution

L’Hôpital applies ONLY when the ratio is in an indeterminate form 0/00/0 or /\infty/\infty. If the limit is in a determinate form, do not use L’Hôpital: you get a wrong result. Moreover, if the limit of the derivatives does not exist, the theorem gives no information (but the original limit might still exist).

Example

limx0ex1x\displaystyle\lim_{x\to 0}\frac{e^x-1}{x}. Form 00\frac{0}{0}. We apply it: (ex)(x)=ex1e01=1\dfrac{(e^x)'}{(x)'} = \dfrac{e^x}{1} \to \dfrac{e^0}{1} = 1.

Example — L'Hôpital applied twice

limx0ex1xx2\displaystyle\lim_{x\to 0}\frac{e^x-1-x}{x^2}. Form 00\frac{0}{0}. First application: ex12x\dfrac{e^x-1}{2x}. Still 00\frac{0}{0}. Second application: ex212\dfrac{e^x}{2}\to\dfrac{1}{2}.

Remark

L’Hôpital is very powerful but is not the only method: equivalent functions (seen in the chapter on limits) are often faster. It is best to use L’Hôpital as a “reserve weapon” when equivalent functions are not enough.

Topics: Calculus theorems
Concepts: Derivative · Indeterminate form · L’Hôpital’s theorem
Methods: L’Hôpital limits
Skills: Computing limits
People: Guillaume de l’Hôpital