Limit Proof?

Case1. When x approaches 0 from positive value

x is positive so x^n is positive. And we know -1 ≦ sin(1/x) ≦ 1, so

-(x^n) ≦ x^n*sin(1/x) ≦ x^n.

And when x approaches 0, both of -(x^n) and x^n approach 0.

So x^n*sin(1/x) approaches 0.

Case2. When x approaches 0 from negative value

Case2A. When n is even

x is negative but x^n is positive.

So similar to case1, x^n*sin(1/x) approaches 0.

Case2B. When n is odd

x is negative so x^n is negative. Therefore, -(x^n) is positive.

And we know -1 ≦ sin(1/x) ≦ 1, so

x^n ≦ x^n*sin(1/x) ≦ -(x^n).

And when x approaches 0, both of x^n and -(x^n) approach 0.

So x^n*sin(1/x) approaches 0.

For all cases, x^n*sin(1/x) approaches 0 when x approaches 0.

First I note that if n > 1, then the values of [x^n * sin(1/x)] in any small neighborhood of x = 0 will be smaller than the values of [x * sin(1/x)] in that same neighborhood.  By a "small" neighborhood I mean one with a diameter  less than 1.

Therefore, it will suffice to prove the proposition for the special case [x * sin(1/x)].

This is actually quite easy.  Although sin(1/x) oscillates wildly in the vicinity of x = 0, nevertheless the absolute value of sin(1/x) is never more than 1, so the function [x * sin(1/x)] lies within the "envelope" of the lines y = x and y = -x, specifically in the two regions that contain the x-axis, and excluding the comparable V-shaped regions that contain the y-axis.  So for any epsilon > 0, the value of [x * sin(1/x)] can be confined to a radius of epsilon just by making sure that |x| < epsilon.