TITLE: Finding the limit of $(1-\cos(x))/x$ as $x\to 0$ with squeeze theorem
QUESTION [7 upvotes]: How do I find:
$$ \lim_{x\to0}\frac{1-\cos(x)}{x} $$
Using the squeeze theorem.  Particularly, how would I arrive at its bounding functions? 
If possible, please try not to use derivatives.

REPLY [2 votes]: For any $x$ such that $0 < |x| \leq \pi/2$, 
$$
0 \le \bigg|\frac{{1 - \cos x}}{x}\bigg| = \frac{{1 - \cos |x|}}{{|x|}} = \frac{{\int_0^{|x|} {\sin t \,dt} }}{{|x|}} \le \frac{{\int_0^{|x|} {\sin |x| \, dt} }}{{|x|}} = \sin |x|.
$$