Squeeze Theorem

[mikky_05v]

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

 

[BigGlenntheHeavy]

\(\displaystyle The \ trick \ here \ is \ sin(?) \ or \ cos(?) \ is \ \ge \ -1 \ and \ \le \ 1.\)

\(\displaystyle For \ example, \ -1\le \ sin(1/x) \ \le \ 1, \ now \ multiply \ through \ by \ x^2 \ and \ you \ get:\)

\(\displaystyle -x^2 \ \le \ x^2sin(1/x) \ \le \ x^2.\)

\(\displaystyle Now, \ taking \ the \ limits \ of \ each \ term \ respectivly, \ gives: \ \lim_{x\to0}-x^2 \ \le\lim_{x\to0}x^2sin(1/x) \ \le \ \lim_{x\to0}x^2.\)

\(\displaystyle This \ gives \ 0 \ \le \ \lim_{x\to0}x^2sin(1/x) \ \le \ 0.\)

\(\displaystyle Therefore, \ only \ one \ way \ to \ go \as \ x^2sin(1/x) \ is \ "squeezed" \ between \ two \ 0s.\)

\(\displaystyle Ergo, \ \lim_{x\to0}x^2sin(1/x) \ = \ 0, \ QED.\)

 

[mmm4444bot]

lim xsin 1/x

lim x^2 sin 1/x

lim x^2 cos 1/x

None of these is a limit statement; they're all missing the value toward which x approaches.

EG:

lim [x -> 5] x sin(1/x)

This limit statement reads, "The limit as x approaches 5 of the product x times sin(1/x)."

I don't remeber the trig functions

You had better review; trigonometry is a major prerequisite of calculus.