For which real values alpha does lim {x -> 0+} x^alpha sin(1/x) exist?

It is easy to show using the epsilon - delta definition below that this limit exists for all real alpha >= 1. In fact the limit is zero in this case. The case alpha equals zero is also quite simple and the limit does not exist. Consider the two sequences

a_n = 2/((4n + 1)pi) and b_n = 2/((4n + 3)pi)

These go to zero through positive values as n --> infinity and yet f(a_n) = 1, however f(b_n) = -1, so this limit cannot exist. I AM INTERESTED IN THE CASE alpha < 0 in which case the limit diverges, but this must be shown rigorously...Thanks!

Definition [Right Hand Limit] Let I be a nonempty interval with a as its left endpoint. A function f:I --> R is said to converge to L as x approaches a from the right if for every epsilon > 0 there is a delta > 0 such that

a < x < a + delta implies |f(x) - L| < epsilon

Solution Preview

The only comment that I can give here is that we know:

If lim[f(x)]=L and lim[g(x)]=M then lim[f(x)g(x)]=LM ; (x->a)

Now, we see that:

Because sin is a bounded function and no matter what x is -1<=sin(x)<=1, then lim(sin(1/x)) ; x->0 is for sure bounded, ...

Solution Summary

A limit is shown to diverge using a rigorous proof.

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