# Limit of: $\lim_{n\to\infty}n\cdot(\sqrt[n]{2011}-1)$ [duplicate]

$\lim_{n\to\infty}n\cdot(\sqrt[n]{2011}-1)$

I rewrote it like this: $\lim_{n \to \infty }\frac{2011^{\frac{1}{n}}-1}{\frac{1}{n}}$, but don't know how to proceed.

Any ideass? Thanks!

## marked as duplicate by hardmath, Davide Giraudo, TooTone, egreg, Nick PetersonApr 4 '14 at 14:43

Set $\displaystyle\frac1n=h,$
finite for $a>0$
$$\lim_{n\to\infty}n\left(\sqrt[n]a -1\right)=\lim_{h\to0}\frac{a^h-1}h$$
Now we know $\displaystyle\lim_{y\to0}\dfrac{e^y-1}y=1$ and $\displaystyle a^h=(e^{\ln a})^h=e^{h\cdot\ln a}$