lim(x-0) xsin(1/x) use the squeezing theorem

Question

lim(x->0)  xsin(1/x) use the squeezing theorem

parthkohli · Answer

Always, $-1 \le\sin \le 1$. Does that give you a proper hint? :)

tiffanymak1996 · Answer

i don't quite understand...

parthkohli · Answer

Have you ever heard of the Squeeze Theorem?(It's also known as 'Sandwich' or 'Pinching' Theorem).

tiffanymak1996 · Answer

yeah, that's the hint given in the question.

parthkohli · Answer

tiffanymak1996 · Answer

wait, i think i know what you mean.

parthkohli · Answer

Great!

parthkohli · Answer

Can you make an inequality to start with?

lgbasallote · Answer

im still wondering how @ParthKohli knows the squeeze

parthkohli · Answer

Also to remember:

$-1 \le \sin x \le 1 \Longrightarrow {-1 \le \sin{1 \over x} \le 1}$

parthkohli · Answer

@lgbasallote I started Calculus a week ago.

parthkohli · Answer

Heh - no need to worry! This is not a copied answer!

lgbasallote · Answer

but this is not "just calculus"

parthkohli · Answer

This is a part of limits, and I am studying Calculus on Paul's Notes.

tiffanymak1996 · Answer

the answer's 0 right?

tiffanymak1996 · Answer

what about lim(x->0+) (lnx)/(cotx)