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I should prove using the limit definition that $\lim_{x \rightarrow 0} \, x^{1/3}\cos(1/x) = 0.$ I have a problem because the second function is much too complex, so I think I need transformation. And what form this function could have in case I will transform it?

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    $-1\le|\cos(1/x)|\le 1$.2011-11-25

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You can solve this problem with the Squeeze Theorem.

First, notice that $-1 \leq \cos(1/x) \leq 1$ (the cosine graph never goes beyond these bounds, no matter what you put inside as the argument).

Multiplying through by $x^{1/3}$, we get $ -x^{1/3} \leq x^{1/3}\cos(1/x) \leq x^{1/3}. $

Now, the Squeeze Theorem says $ \lim_{x \rightarrow 0} (-x^{1/3}) \leq \lim_{x \rightarrow 0} \, x^{1/3}\cos(1/x) \leq \lim_{x \rightarrow 0} \, x^{1/3}, $ so we investigate the left- and right-most limits.

Since, $x^{1/3}$ is continuous on $[0,\infty)$, $ \lim_{x \rightarrow 0} \, x^{1/3} = \lim_{x \rightarrow 0} (-x^{1/3}) = 0. $

Finally, we have $0 \leq \lim_{x \rightarrow 0} \, x^{1/3}\cos(1/x) \leq 0,$ which forces us to conclude $ \lim_{x \rightarrow 0} \, x^{1/3}\cos(1/x) = 0. $

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    I just now noticed that your question asked to prove "from the limit definition", by which I assume you mean the $\epsilon-\delta$ definition. In that case, bounding $|\cos(1/x)|$ by $1$ is still the way to go, but you'll need to show more details than I have here.2011-11-25