4
$\begingroup$

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

$\endgroup$
4
  • 1
    $\begingroup$ 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 $x\to 0$. $\endgroup$ Sep 28, 2012 at 18:38
  • $\begingroup$ Even if I took another bound in the left, I still have infinity as a result of the limit in the right. $\endgroup$
    – ChairOTP
    Sep 28, 2012 at 18:44
  • $\begingroup$ 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 $\cos^2 x+\sin^2 x = 1$. $\endgroup$ Sep 28, 2012 at 18:46
  • $\begingroup$ 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. $\endgroup$
    – ChairOTP
    Sep 28, 2012 at 19:19

2 Answers 2

2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
11
  • $\begingroup$ As in multiplying by $1+\cos x \over {1+\cos x}$ ? $\endgroup$
    – ChairOTP
    Sep 28, 2012 at 18:54
  • $\begingroup$ That should work, yes. $\endgroup$
    – EuYu
    Sep 28, 2012 at 18:59
  • $\begingroup$ 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? $\endgroup$
    – ChairOTP
    Sep 28, 2012 at 19:06
  • $\begingroup$ 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. $\endgroup$
    – EuYu
    Sep 28, 2012 at 19:09
  • $\begingroup$ 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? $\endgroup$
    – ChairOTP
    Sep 28, 2012 at 19:14

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .