# Squeeze Theorem

• Sep 7th 2010, 05:13 PM
shmiks
Squeeze Theorem
During class today we started covering some stuff that had me totally lost. We were covering The squeeze theorem and I was doing really good up until
lim xsin 1/x
I haven't the faintest idea what he did or how to repeat it. Can anyone give me a thorough explanation of this? The guy speaks terrible english and after his very brief run through of it simply told us to do
lim x^2 sin 1/x
lim x^2 cos 1/x
at home and know how to do them for tomorrows quiz.
So any advise on where to begin with those would be awesome. I think i'm mostly lost bc I don't remeber the trig functions i need. (like tanx=sinx/cosx). But I may be way off. Help
• Sep 7th 2010, 05:36 PM
skeeter
note from the graph that $-|x| \le x \sin\left(\frac{1}{x}\right) \le |x|$

so ... what can you say about the limit of the function in question as $x \to 0$ ?
• Sep 7th 2010, 05:48 PM
shmiks
I don't understand where you come up with the x and -x at all. I got to the graph part and don't know what i'm looking for from there. And that doesn't help me understand what to do with the other 2 problems.
x^2sin1/x
x^2cos1/x
which are my main concern.
I also understand what the answer of xsin1/x is. i just need to know how to get there.
• Sep 8th 2010, 04:00 PM
lvleph
Are you taking the limit as x approaches 0?

First, we know that $|\sin \frac{1}{x}| \le 1$ thus it should be clear that $0 \le |x \sin \frac{1}{x}| \le |x| \Leftrightarrow -|x| \le x \sin \frac{1}{x} \le |x|$

Now, $\lim_{x\to 0} |x| = 0$. From here apply the squeeze theorem.