proof of special trig limits I'm trying to prove a special trig limit, which is...
$$\lim_{x \rightarrow 0} \frac{1 - \cos{x}}{x}=0$$
So far, this is what I have (and I'll explain where I'm confused)
Using the squeeze theorem,
$h(x) \leq f(x) \leq g(x)$
$$-x^2 + 1 \leq \cos{x} \leq 1 $$
$$-x^2 + 1 - 1 \leq \cos x - 1 \leq 1 - 1$$
$$-x^2 \leq \cos{x} - 1 \leq 0$$
$$0 \leq 1 - \cos{x} \leq x^2 $$
$$0 \leq \frac{1- \cos{x}}{x} \leq x$$
Since limit of $0$ and $x$ equals zero (as $x$ approaches zero), so does $\displaystyle{\frac{1-\cos{x}}{x}}$.
My first confusion, is when I try to graph the last line as separate functions
In quadrant $3$ and $4$, it holds up. However, in quadrant $1$ and $2$, it becomes 
$\displaystyle{x \leq \frac{1-\cos{x}}{x} \leq 0}$.
I'm not sure if this is allowed in squeeze theorem, but I'm a tad bit confused.
I have another guess as to why the end result is incorrect because $-x^2 + 1 \leq \cos{x} \leq 1$  are not the correct "sandwich" functions. $f(x) = \cos{x}$ touches $g(x) = 1$ in more than one spot. Every diagram of squeeze theorem I've seen, the sandwich functions only touch $f(x)$ at one spot. Is this a criteria I'm unaware of for picking $h(x)$ and $g(x)$?
PS, I know I could have used $h(x) = -x^2 + 1$ and $g(x) = x^2 + 1$, but I'd still like to know what I did wrong up top, please.
Thanks for any helps, guys/gals/automatons

I would recommend that you use L'Hospital's Rule. That's probably easier, since you can get rid of the denominator instantly.
$$\lim_{x \to 0} \frac{1-\cos x}{x} \rightarrow \lim_{x \to 0} \frac{\sin x}{1} = 0$$
