Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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?

Answer 4

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