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Evaluate the limit $$\lim_{n\to+\infty}(\sqrt[n]{n}-1)^n$$

I know the limit is 0 by looking at the graph of the function, but how can I algebraically show that that is the limit?

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Showing $\sqrt[n]{n}-1 \to 0$ would be a good start (actually, you need less). –  Daniel Fischer Nov 21 '13 at 14:45

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$$a_n:=\left(\sqrt[n]n-1\right)^n\implies\log a_n=n\log\left(\sqrt[n]n-1\right)=-\infty\implies$$

$$\lim_{n\to\infty} a_n=\lim_{n\to\infty}e^{\log a_n}=0$$

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