Hey guys having trouble solving this . Use l’Hopital’s rule to find:

$\displaystyle \lim_{x\to0}\frac{cosx-1}{x^2}$

any help much appreciated

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- Mar 27th 2009, 08:59 PMjvignaciol’Hopital’s Rule Help..
Hey guys having trouble solving this . Use l’Hopital’s rule to find:

$\displaystyle \lim_{x\to0}\frac{cosx-1}{x^2}$

any help much appreciated - Mar 27th 2009, 09:19 PMADARSH
Differentiate numerator and denominator

$\displaystyle \frac{d~}{dx} (cos(x) - 1 ) = -sin(x) $

$\displaystyle \frac{d}{dx} (x^2) = 2x

$

What is

$\displaystyle (\frac{-1}{2})\lim_{x\to0}\frac{sin(x)}{x}$ ?

You can do the same for finding this limit or just use the formula (which??) - Mar 27th 2009, 09:36 PMjvignacio
- Mar 27th 2009, 09:45 PMADARSH

1)Find derivatives separately

2) Divide the numerator's derivative by denominator's derivative

3)If on putting the limit you get a definite answer than game over!!

4) If not than go to step 1

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I strongly recommend reading examples here

The answer of this question will be -1/2 (after you follow above steps) (Happy)

Adarsh

- Mar 27th 2009, 10:40 PMjvignacio
- Mar 27th 2009, 11:06 PMADARSH
Denominator =D Numerator =N

Derivative of Numerator $\displaystyle = \frac{d~}{dx} (cos(x) - 1 ) = -sin(x)$...(Also called N'(x))

Derivative of Denominator $\displaystyle = \frac{d}{dx} (x^2) = 2x$...(Also called D'(x) )

According to the l'Hopital's rule

http://upload.wikimedia.org/math/7/0...97ae0096aa.png

So using this

Your Limit now becomes

http://www.mathhelpforum.com/math-he...daea34ad-1.gif

Once again applying the same rule this limit becomes

$\displaystyle (\frac{-1}{2}) lim_{x\to 0} cos(x)$

Now put the limit

$\displaystyle \implies (\frac{-1}{2}) \times (1)$