help with a limit

**SimonK** · March 20th 2010, 08:00 PM

Hi
I need a bit of help here....not sure how to start with
lim x->0 (cos(^2)x - cosx)/x

if i take cosx out [cosx(cosx - 1)]/x
doesn't that give me cosx/x out the front which is undefined?....or am I getting all turned around?

Sorry about how messy this is....I'm still learning how to use latex

Thanks every one

**Prove It** · March 20th 2010, 08:14 PM

Originally Posted by SimonK

Hi
I need a bit of help here....not sure how to start with
lim x->0 (cos(^2)x - cosx)/x

if i take cosx out [cosx(cosx - 1)]/x
doesn't that give me cosx/x out the front which is undefined?....or am I getting all turned around?

Sorry about how messy this is....I'm still learning how to use latex

Thanks every one

Since this tends to $\frac{0}{0}$ you can use L'Hospital's Rule.

$\lim_{x \to 0}\frac{\cos^2{x} - \cos{x}}{x}$

$= \lim_{x \to 0}\frac{\frac{d}{dx}(\cos^2{x} - \cos{x})}{\frac{d}{dx}(x)}$

$= \lim_{x \to 0}\frac{-2\sin{x}\cos{x} + \sin{x}}{1}$

$= \lim_{x \to 0}\,(\sin{x} - 2\sin{x}\cos{x})$

$= 0$ .

**SimonK** · March 20th 2010, 08:21 PM

Thanks very much

**Pulock2009** · March 20th 2010, 10:54 PM

Originally Posted by SimonK

Thanks very much

cosx(cosx-1)/x
=cosx(2sin^2(x/2))x/x^2
=sin part becomes 1 and cos part becomes 0
=0 hence the answer.

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