[SOLVED] another trig limit question

**dannyboycurtis** · October 19th 2009, 09:26 PM

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?

**mr fantastic** · October 19th 2009, 09:36 PM

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?

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

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

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

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