Question

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

Answer 1

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

Answer 2

i don't quite understand...

Answer 3

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

Answer 4

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

Answer 5

So you haven't heard of it. Hmm.
I recommend you to see this link.
http://tutorial.math.lamar.edu/Classes/CalcI/ComputingLimits.aspx

Answer 6

wait, i think i know what you mean.

Answer 7

Great!

Answer 8

Can you make an inequality to start with?

Answer 9

im still wondering how @ParthKohli knows the squeeze

Answer 10

Also to remember:

\(-1 \le \sin x \le 1 \Longrightarrow {-1 \le \sin{1 \over x} \le 1}\)

Answer 11

@lgbasallote I started Calculus a week ago.

Answer 12

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

Answer 13

but this is not "just calculus"

Answer 14

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

Answer 15

the answer's 0 right?

Answer 16

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