Limit Proof?
Show that the limit of [x^n • sin(1/x)] as x tends to zero is 0, where n is a positive integer.
Show that the limit of [x^n • sin(1/x)] as x tends to zero is 0, where n is a positive integer.
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Favorite Answer
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.
az_lender
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.