How I can understand this:

Lim(x*sin(1/x)) when x goes to 0 equals to 0
? Thanks!

Question

How I can understand this: 

 Lim(x*sin(1/x)) when x goes to 0 equals to 0
? Thanks!

anonymous · Answer

sine fluctuates very quickly, because as $x\to 0$ you have $\frac{1}{x}\to \infty$ 
but sine is stuck between -1 and 1, so $x\sin(\frac{1}{x})$ is stuck between $-x$ and $x$

anonymous · Answer

and as $x\to0$ clearly $-x\to 0$ as well, so the whole thing goes to zero 
here is a picture 
www.wolframalpha.com/input/?i=xsin(1%2Fx)

you can  sort of see that the function $x\sin(\frac{1}{x})$ is bounded by the two lines $y=x$ and $y=-x$

jpsmarinho · Answer

This is the confrontation theorem, right?

anonymous · Answer

i have never heard of the that theorem, but maybe.
sometimes i have heard this called the "squeeze" theorem, but for all i know they are the same

jpsmarinho · Answer

I'm Brazilian, here it's called confrontation, but I've heard squeeze. Thanks!!