Proving limit of $\sqrt{n} (\sqrt[n]{n} -1)$ is $0$ as $n \to \infty$. I managed to prove that limit to $\sqrt[n]{n} -1$ tends to $0$, but couldn't go anywhere from there. Could somebody help?
We have
\begin{align}
\ln(\sqrt{n}(\sqrt[n]{n}-1)) &amp;= \ln(\sqrt{n}) + \ln(\exp{\frac{\ln(n)}{n} -1})\\
&amp;= \ln(\sqrt{n}) + \ln(\frac{\ln(n)}{n} + o(\frac{\ln(n)}{n}))\\
&amp;= \ln(\frac{\ln(n)}{\sqrt{n}} + o(\frac{\ln(n)}{\sqrt{n}}) \rightarrow - \infty
\end{align}
Thus
$$
\sqrt{n}(\sqrt[n]{n}-1) \rightarrow 0
$$
as $n \rightarrow + \infty$.
Using l'Hôpital's rule, the derivative of $f(n)=\sqrt{n} (\sqrt[n]{n} -1)$ is:

$$
f'(n)=n^{1/n + 1/2} \frac{1-\log(n)}{n^2} + \frac{\sqrt[n]{n} -1}{2 \sqrt{n}}.
$$

Now taking $\displaystyle  \lim_{n\to\infty}$ of the above expression, we get $\displaystyle 0 + \lim_{n\to\infty}  \frac{\sqrt[n]{n} -1}{2 \sqrt{n}}$, and we know that $\displaystyle \lim_{n\to\infty} \sqrt[n]{n} = 1$, which yields the desired result.