Find the limit of $n \ (\sqrt[n]{n} - 1) $ 
Find the limit of $n \ (\sqrt[n]{n} - 1) $.

Wolfram says it is $+\infty$.
Obviously $n \rightarrow +\infty$ and $\sqrt[n]{n} - 1\rightarrow 0$.
Any hint?
 A: With Taylor approximations, using the facts that (i) $e^x = 1+x +o(x)$ when $x\to 0$, (ii) $\sqrt[n]{n}=e^{\frac{\ln n}{n}}$, and (iii) ${\frac{\ln n}{n}}\xrightarrow[n\to\infty]{}0$:
(See below for a full, detailed derivation; stop here if you just wanted a hint.)


$$\begin{align}n\left(\sqrt[n]{n}-1\right) &= n\left(e^{\frac{\ln n}{n}}-1\right)= n\left(1+\frac{\ln n}{n} + o\left(\frac{\ln n}{n}\right)-1\right) \\&= n\left(\frac{\ln n}{n} + o\left(\frac{\ln n}{n}\right)\right)= \ln n + o(\ln n)\end{align}$$and since $\ln n \xrightarrow[n\to\infty]{} \infty$ you can conclude by comparison.

A: Hint 
$$
 n \ (\sqrt[n]{n} - 1)={e^{(\ln n)/n}-1\over (\ln n)/n}\ln n.
$$
A: $$\lim_{n\to\infty}\frac{(n^\frac1n-1)}{\frac1n}$$
Using Hopital
$$=\lim_{n\to\infty}\frac{n^{\frac1n}\frac1{n^2}\times (-\ln n +1)}{\frac{-1}{n^2}}$$
$$=\lim_{n\to\infty}n^{\frac1n}(\ln n -1)$$
A: $$n(\sqrt[n] n-1)= \frac {\sqrt[n] n-1}{\frac{1}{n}}$$ Then applying L'Hôpital's rule, one has (after simplification in numerator and denominator of the factor $\frac{1}{n^2}$ which  I can't arrive to write in Tex Commands) $$\frac{-n^{1/n}(\ln n-1)}{-1}$$ and this expression clearly tends to $+\infty$ when $n\to\infty$
