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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

ChairOTP
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    You didn't squeeze hard enough, i.e. your bounds $-1$ and $1$ are too trivial (or at least $-1$ is). You need something that gets "tight" when $x\to 0$. – Hagen von Eitzen Sep 28 '12 at 18:38
  • Even if I took another bound in the left, I still have infinity as a result of the limit in the right. – ChairOTP Sep 28 '12 at 18:44
  • The $\infty$ on the right is a consequence of using merely $-1\le\cos x$ (though this got turned to $-\cos x\le 1$ inbetween). You need som estimate $f(x)\le \cos x$ with the property that $f(x)\to0$ as $x\to 0$. Hint: make use of $\cos^2 x+\sin^2 x = 1$. – Hagen von Eitzen Sep 28 '12 at 18:46
  • 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. – ChairOTP Sep 28 '12 at 19:19

2 Answers2

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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.

kalvotom
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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.

Michael Hardy
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EuYu
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  • As in multiplying by $1+\cos x \over {1+\cos x}$ ? – ChairOTP Sep 28 '12 at 18:54
  • That should work, yes. – EuYu Sep 28 '12 at 18:59
  • I split it into two limits, the one you suggested and $1\over {1+\cos x}$. Can I evaluate the second one before using the Squeeze Theorem? – ChairOTP Sep 28 '12 at 19:06
  • In practice it wouldn't make a difference if you evaluated it right now. To be perfectly formal, I would probably wait until the end step when you take $x$ to $0$ for the squeeze. – EuYu Sep 28 '12 at 19:09
  • Correct me if I'm wrong (which I think I am) but then I would have to evaluate $(-1\le {{\sin x}^2 \over {x^2}} \le 1)\times (-1\le \cos x \le 1)$ The second one until I get to $1\over {1+\cos x}$ so I can squeeze both? – ChairOTP Sep 28 '12 at 19:14
  • The problem of squeezing the function $\frac{sin(x)}{x}$ is very well known. The only squeeze based solution that I'm aware of will rely on geometric methods. How are you multiplying a set of inequalities? – EuYu Sep 28 '12 at 19:17
  • No, the problem is not squeezing $\frac{sin(x)}{x}$, I've done it already, my question was that if it's OK to split them the way I did, and then squeeze both parts separately. – ChairOTP Sep 28 '12 at 19:21
  • $1+cosx$ simply goes to $2$. There's no squeezing involved... – EuYu Sep 28 '12 at 19:23
  • Look I have this then:

    $\lim_{x\to 0} {\sin x^2\over x^2} \times \lim_{x\to 0} {1 \over {1+\cos x}}$ Then: $\frac12 \times \lim_{x\to 0} {\sin x^2\over x^2}$ and then I just do the Squeeze Theorem like this: $\frac12 \times (-1\le {{\sin x}^2 \over {x^2}} \le 1)$, is that correct?

     – ChairOTP Sep 28 '12 at 19:24
  • Yes, that seems fine. Just a note, the \frac{sin^2(x)}{x^2} is squeezed to $1$. All you wrote is an inequality, not a squeeze. Assuming you've done the squeeze properly (which you did not write), then it's fine. – EuYu Sep 28 '12 at 19:31
  • Oh yes I know, I was just asking if that's the previous step before I use the squeeze theorem in that inequality. – ChairOTP Sep 29 '12 at 02:40