<s>[INST] Evaluate a limit by using squeeze theorem We're supposed to use the Squeeze Theorem to prove that 
$$\lim_{x\to 0} {1-\cos x\over x^2} = \frac12$$
I tried this:
$$-1\le \cos x \le 1$$ 
$$-1\le -\cos x \le 1$$
$$0\le 1-\cos x \le 2$$
$$0\le {1-\cos x\over x^2} \le {2\over x^2}$$
Then using limits we have:
$$\lim_{x\to 0}0\le \lim_{x\to 0} {1-\cos x\over x^2} \le \lim_{x\to 0}{2\over x^2}$$
And for obvious reasons the first limit is $\Bbb {0}$, and the third limit is $\Bbb \infty$
What do I do now? Or what am I doing wrong?
Thanks in advance
 [/INST] This might be an overkill, but according to the Taylor theorem, for any nonzero $x$ you can find $\xi_x$ between zero and $x$ in such a way that
$$
  \cos x = 1 - \frac{x^2}{2} + \frac{1}{4!} \cos(\xi_x) \cdot x^4.
$$
Thus, shuffling those terms around, you would get
$$
  \frac{1}{2} - \frac{x^2}{4!} \leq  \frac{1 - \cos x}{x^2} = \frac{1}{2} - \frac{x^2}{4!} \cos(\xi_x) \leq \frac{1}{2} + \frac{x^2}{4!}, \quad x \neq 0.
$$
Obviously
$$
  \lim_{x\to 0} \frac{1}{2} \pm \frac{x^2}{4!} = \frac{1}{2}
$$
and you are done.
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