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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

  • 0
    So you're suggesting I could take two different bounds starting off $\cos x$? Or should I start off from something else, because I've tried already to change bounds but it will lead me to $0$ in one side.2012-09-28

2 Answers 2

1

Your bounds do not seem tight enough. If you know how to squeeze $\frac{\sin(x)}{x}$ then one possible solution would be to reduce your limit into $\frac{\sin^2(x)}{x^2}$ and to squeeze that.

  • 0
    Oh yes I know, I was just asking if that's the previous step before I use the squeeze theorem in that inequality.2012-09-29
2

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.