Advanced Calc/Analysis: Delta Epsilon proof

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Homework Statement


Using the definition of |x-a|<delta implies |f(x) - L|<epsilon, prove that lim x->0 x^n*sin(1/x) holds for all n belonging to natural numbers.

Homework Equations


Definition of a limit

The Attempt at a Solution


Ok, so when I see "prove for all n belonging to natural numbers" I immediately think induction. So this is what I have done so far. For n=1 lim x->0 x^n*sin(1/x) is true; the limit is zero. Now I will assume lim x->0 x^n*sin(1/x) for some n is true, then I need to show that n+1 is also true. So I start using the definition of the limit and don't know what my L should be and how to use induction along with this definition of a limit. Please help and thank you in advanced.
 
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\mbox{If you can prove that }\lim_{x\to 0}x^{n}sin\left(\frac{1}{x}\right)\mbox{ is bounded by two functions whose limit is L, then you&#039;ve proven that the limit exists.}
\mbox{(You don&#039;t have to invoke the Squeeze Theorem.)}

\mbox{Start with }-1\leq \sin\left(\frac{1}{x}\right)\leq 1\mbox{. What can you say about }x^{n}sin\left(\frac{1}{x}\right)\mbox{ ?}

\mbox{It isn&#039;t completely necessary, but it might help you see what L to use if you make the substitution x=1/t, then }

\lim_{x\to 0}x^{n}sin\left(\frac{1}{x}\right) = \lim_{t\to\infty}\left(\frac{1}{t}\right)^{n}sin\left(t\right)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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