a limits inquiry?

Question

find the limit as x-->0 of [xsin(1/x)] If the limit does exist, find it; then, given an arbitrary epsilon, find a delta (as a function of epsilon) such that |f(x) - L| < Epsilon whenever |x-a|
This is the only problem that's giving me trouble on my workshop. I need help.

Answer 1

For the function x sin(1/x), the dominant term is the x as the sine function stays in between -1 and 1. Indeed, I don't think that sin(1/x) has a limit as x ->0, but is a finite value, so using the product rule of limits where lim f(x)*g(x) = lim f(x) * lim g(x), then since x ->0, so does the function given.

Answer 2

No hell no, you are guys are just doing wayyyyyyyyyyy too much work on this.

The answer is that the limit does exist and it is zero.

How to prove it, use the squeeze/sandwich theorem.
Since the sine function is always bounded above by 1 and bounded below by -1, x*sin(1/x) is bounded above by abs(x) and bounded below by -abs(x) and since both of these go to zero, x*sin(1/x) has to go to zero as well.

As for the epsilon and delta proof, I say let delta=epsilon.

Answer 3

The limit is 0.

For every x≠0, -1≤sin(1/x)≤1. Therefore |sin(1/x)|≤1 and |xsin(1/x)|≤|x|.

let ε>0 be arbitrary. Choose δ=ε. For every x such that δ>|x-0|=|x|, |f(x)-0|=|xsin(1/x)|≤|x|<δ=ε