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When can you use L'Hopital's rule

L'Hôpital's rule applies when a limit produces the indeterminate forms 0/00/0 or /\infty/\infty, the numerator and denominator are differentiable on an open interval around the point (except possibly at the point), and the limit of their derivatives exists (finite or infinite). If those conditions hold, limxaf(x)g(x)=limxaf(x)g(x)\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)} provided the latter limit exists.

Calculus · Limits


L'Hôpital's rule turns a hard indeterminate limit into a usually simpler one by differentiating the top and bottom. It is a shortcut, not a universal tool, so checking the hypotheses prevents wasted work.

Key Conditions

You must check all of these:

  • Both ff and gg are differentiable on an interval around aa (except possibly at aa)
  • limxaf(x)=limxag(x)=0\lim_{x\to a}f(x)=\lim_{x\to a}g(x)=0 or both are infinite, giving 0/0 or /\infty/\infty
  • g(x)0g'(x)\neq0 on that interval
  • The limit limxaf(x)g(x)\lim_{x\to a}\frac{f'(x)}{g'(x)} exists (or is ±\pm\infty )

Apply the rule step‑by‑step:

  1. 1Confirm the original limit is 0/0 or /\infty/\infty
  2. 2Differentiate numerator and denominator separately
  3. 3Take the limit of the new fraction
  4. 4If the new limit is still indeterminate, repeat the process

Typical indeterminate forms and whether L'Hôpital applies

FormApplicable?
0/0Yes
/\infty/\inftyYes
0\cdot\inftyNo (rewrite)
1^{\infty}No (rewrite)

Concrete Example

Evaluate limx0sinxx\displaystyle\lim_{x\to0}\frac{\sin x}{x}. Direct substitution gives 0/0, so differentiate: ddxsinx=cosx\frac{d}{dx}\sin x=\cos x and ddxx=1\frac{d}{dx}x=1. The new limit is limx0cosx1=cos0=1\lim_{x\to0}\frac{\cos x}{1}=\cos0=1, so the original limit equals 1.

Check yourself

Which of the following is a required condition for L'Hôpital's rule?

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