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l'Hôpital's rule
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Definition/Summary
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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. |
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Equations
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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. |
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Recent forum threads on l'Hôpital's rule
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Breakdown
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Mathematics
> Calculus/Analysis
>> Functions
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Images
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Extended explanation
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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. |
Commentary
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Eidos @ 02:12 PM May22-08
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