Use 1' Hopital's Rule t find the limit. SHOW WORK Lim cos x-1 /2x(squared) is 0/0 form.

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- Dec 18th 2008, 06:53 AM #1

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- Dec 18th 2008, 07:17 AM #2

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Hi there,

Whenever you have a limit of indeterminate form of type 0/0

just differentiate the top and the bottom separately.

$\displaystyle

\frac{d}{dx}(Cos(x)-1)= -Sin(x)

$

$\displaystyle

\frac{d}{dx}(2x\sqrt{2})=\sqrt{2}\frac{d}{dx}(2x)= 2 \sqrt{2}

$

Now you take the limit of f'(x)/g'(x)

$\displaystyle

\lim_{x \rightarrow 0}\frac{-sin(x)}{2\sqrt{2}}=0

$

NOTE: Here we differentiated only once, but sometimes you have to differentiate multiple times in order to get the limit.

- Dec 18th 2008, 11:40 PM #3

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