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What is l'Hôpital's rule

  1. Jul 23, 2014 #1

    L'Hôpital's (or l'Hospital's) rule is a method for finding the limit of a function with an indeterminate form.


    If the expression

    [tex]\frac{\lim_{x \rightarrow a} f(x)}{\lim_{x \rightarrow a} g(x)}[/tex]

    has the form [itex]0/0[/itex] or [itex]\infty / \infty[/itex], then l'Hôpital's rule states that

    [tex]\lim_{x \rightarrow a} \frac{f(x)}{g(x)}
    = \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)}[/tex]

    provided that that second limit exists.

    Extended explanation


    [tex]1.~~\lim_{x\rightarrow 0}\frac{\sin x}{x}\,=\,\lim_{x\rightarrow 0}\frac{\cos x}{1}\,=\,1[/tex]

    [tex]2.~~\lim_{x\rightarrow 0}\frac{e^x-1}{x}\,=\,\lim_{x\rightarrow 0}\frac{e^x}{1}\,=\,1[/tex]

    The rule can be applied more than once:

    If after one application, the ratio is still of the form [itex]0/0[/itex] or [itex]\infty / \infty[/itex], then the rule may be applied again (and as many times as are needed to produce a limit):

    [tex]3.~~\lim_{x\rightarrow 0}\frac{e^x\,-\,x\,-1}{\frac{1}{2}x^2}\,=\,\lim_{x\rightarrow 0}\frac{e^x\,-\,1}{x}\,=\,\lim_{x\rightarrow 0}\frac{e^x}{1}\,=\,1[/tex]

    Example of the rule not helping:

    It is possible that the limit of the ratio of the derivatives does not exist, even though the limit of the original ratio does:

    [tex]\lim_{x\rightarrow \infty}\frac{x\ +\ \sin x}{x}\ =\ 1[/tex]

    but [tex]\lim_{x\rightarrow \infty}\frac{1 +\ \cos x}{1}[/tex] does not exist.

    * This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!
  2. jcsd
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