Thread: [SOLVED] another trig limit question

1. [SOLVED] another trig limit question

This one is throwing me for a loop:
$\lim_{x\to 0}\frac{1-cosx}{x}=0$
how do I show this using the fact that $\lim_{x\to 0}\frac{sinx}{x}=1$?
I realize this is more a question of manipulating the function using trig identities, but I really cant see it.
Any help?

2. Originally Posted by dannyboycurtis
This one is throwing me for a loop:
$\lim_{x\to 0}\frac{1-cosx}{x}=0$
how do I show this using the fact that $\lim_{x\to 0}\frac{sinx}{x}=1$?
I realize this is more a question of manipulating the function using trig identities, but I really cant see it.
Any help?
$\lim_{x\to 0}\frac{1-\cos x}{x}= \lim_{x\to 0}\frac{(1- \cos x)(1 + \cos x)}{x(1 + \cos x)}$

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

$= \lim_{x\to 0}\frac{\sin x}{x} \cdot \lim_{x\to 0}\frac{\sin x}{(1 + \cos x)}$.