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

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

$$a_n:=\left(\sqrt[n]n-1\right)^n\implies\log a_n=n\log\left(\sqrt[n]n-1\right)=-\infty\implies$$
$$\lim_{n\to\infty} a_n=\lim_{n\to\infty}e^{\log a_n}=0$$
Since, $\frac{\log(x)}x\le\frac1e$, we have that \begin{align} \sqrt[n]{n}-1 &\le e^{1/e}-1\\ &\lt1 \end{align} Therefore, \begin{align} \lim_{n\to\infty}\left(\sqrt[n]{n}-1\right)^n &\le\lim_{n\to\infty}\left(e^{1/e}-1\right)^n\\[3pt] &=0 \end{align}