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

Equations

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

Examples:

[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.

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# What is l'Hôpital's rule

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