Small Proof

Murk

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Feb 23, 2010
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\(\displaystyle n \in \mathbb{N}\), \(\displaystyle sin(1/x)/x^n \to 0\), as \(\displaystyle x\to \infty\). Only thing I know is \(\displaystyle sin(1/x)/x^n\leq1/x^n\). How should I even begin to start to prove this statement?
 
I think I got it now but I am not sure if this can be done. Can we simply say since \(\displaystyle sin(1/x) \cdot 1/x^n\) and since \(\displaystyle sin(1/x) \to 0\) and \(\displaystyle 1/x^n \to 0\) as \(\displaystyle x \to \infty\) implies that \(\displaystyle sin(1/x)/x^n \to 0\) as \(\displaystyle x \to \infty\). Can I do this in this problem?
 
Yes.Your first argument would have worked too. By the squeeze theorem,

for x-> infinity (might as well assume x>1), 0 <= sin(1/x) <= 1

Hence 0 <= lim sin(1/x)/x^n <= lim 1/x^n = 0
 
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