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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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$-1\le|\cos(1/x)|\le 1$. –  David Mitra Nov 25 '11 at 15:18

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up vote 3 down vote accepted

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. –  Austin Mohr Nov 25 '11 at 15:41

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