finding a limit with trig

• September 30th 2008, 12:54 PM
bene
finding a limit with trig
Question

find the limit or prove it does not exist

limx->0 xsinx/1-cosx

just pluging this into my calculator it looks like the limit is 2 but i can't figure out how to get this... i should mention it looks like there is no slope there, i know that means that there is no derivative there but i forgot if that also means the limit is undefined?

thanks
• September 30th 2008, 01:11 PM
icemanfan
Beginning with $\frac{x \sin x}{1 - \cos x}$, multiply by $\frac{1 + \cos x}{1 + \cos x}$ to obtain:

$\frac{x (\sin x)(1 + \cos x)}{1 - \cos^2 x}$

$\frac{x (\sin x)(1 + \cos x)}{\sin^2 x}$

$\frac{x(1 + \cos x)}{\sin x}$

$\frac{x}{\sin x} \cdot {(1 + \cos x)}$

The limit on the left is 1, and the limit on the right is 2. Hence the limit of the product is 2.
• September 30th 2008, 01:15 PM
bene
awesome thank you very much