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l'Hôpital's rule

 Definition/Summary 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 $$\frac{\lim_{x \rightarrow a} f(x)}{\lim_{x \rightarrow a} g(x)}$$ has the form $0/0$ or $\infty / \infty$, then l'Hôpital's rule states that $$\lim_{x \rightarrow a} \frac{f(x)}{g(x)} = \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)}$$ provided that that second limit exists.

 Scientists Guillaume François Antoine, Marquis de l'Hôpital (1661 – 1704), better known as Guillaume de l'Hôpital or Marquis de l'Hôpital. Pronunciation guide: (gee-yome) (frahn-swah) (ahn-twahn) (mahr-kee) (the) (low-pee-thahl)

 Recent forum threads on l'Hôpital's rule

 Breakdown Mathematics > Calculus/Analysis >> Functions

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 Extended explanation Examples: $$1.~~\lim_{x\rightarrow 0}\frac{\sin x}{x}\,=\,\lim_{x\rightarrow 0}\frac{\cos x}{1}\,=\,1$$ $$2.~~\lim_{x\rightarrow 0}\frac{e^x-1}{x}\,=\,\lim_{x\rightarrow 0}\frac{e^x}{1}\,=\,1$$ The rule can be applied more than once: If after one application, the ratio is still of the form $0/0$ or $\infty / \infty$, then the rule may be applied again (and as many times as are needed to produce a limit): $$3.~~\lim_{x\rightarrow 0}\frac{e^x\,-\,x\,-1}{\frac{1}{2}x^2}\,=\,\lim_{x\rightarrow 0}\frac{e^x\,-\,1}{x}\,=\,\lim_{x\rightarrow 0}\frac{e^x}{1}\,=\,1$$ 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: $$\lim_{x\rightarrow \infty}\frac{x\ +\ \sin x}{x}\ =\ 1$$ but $$\lim_{x\rightarrow \infty}\frac{1 +\ \cos x}{1}$$ does not exist.

Commentary

 tiny-tim @ 02:09 PM Jan8-09 Added "provided that that second limit exists", and an example of repeated use.

 @ 01:39 AM May24-08 this rule is actually invented by the teacher of L' HOSPITAL named BERNOULI.

 Ed Aboud @ 04:01 PM May23-08 Ye it should

 Eidos @ 02:12 PM May22-08 Also didn't L'Hopital buy this from Bernoulli?

 Nick89 @ 01:25 PM May21-08 Bah, no latex? What I meant to say was: lim( [e^x - 1]/x ) = lim( [e^x] / 1 ) = e^0 = 1

 Nick89 @ 01:24 PM May21-08 Shouldn't the second example yield 1? [cctex] \lim_{x\rightarrow 0}\frac{e^x-1}{x}=\lim_{x\rightarrow 0}\frac{e^x}{1}=e^0=1 [/cctex]