## amorable 3 years ago Squeeze Theorem Show that -|x| less than equal to x sin (1/x) less than equal to |x| The use the squeeze theorem to show that the limit as approaches 0 of x sin (1/x)=0 Can someone walk me through the squeeze theorem? I don't understand how to solve the problems.

1. LolWolf

Well, to show both cases we simply have to show that $$|x|\ge xsin\left(\frac{1}{x}\right)$$. Since we know, $$\forall x\in \mathbb{R}, |\sin(x)|\le1$$, then, multiplying both sides by $$x$$ gives us: $x|\sin(x)|\le x\implies x\sin(x)\le x$And thus, since $$|x|\ge x$$ $x\sin(x)\le|x|$We know: $-|x|\le x$Which means that (given some more, similar work to the above): $x\sin(x)\ge-|x|$Therefore: $-|x|\le x\sin(x) \le |x|$And, we are done.

2. LolWolf

For the second part, we know: $\lim_{x\to 0}-|x|=0\\ \lim_{x \to 0}|x|=0$And, since: $-|x|\le x\sin\left(\frac{1}{x}\right) \le |x|$We then have that, by the squeeze theorem. $\lim_{x\to 0}\left(x\sin\left(\frac{1}{x}\right)\right)=0$ And I just realized I made a mistake on the last post. Everytime I mention $$x\sin(x)$$ I actually mean $$x\sin\left(\frac{1}{x}\right)$$. My bad.

3. amorable

Thank you! The way you put it makes it easier to understand than the book. Does this work the same for cosine and secant limits?

4. LolWolf

Yes for cosine, as for secant, it's slightly different, but try it yourself, it's more fun, that way. And sure thing.