$\lim\limits_{n\rightarrow\infty} \frac{n}{\log n}(n^{\frac{1}{n}}-1)$ How to find $\lim\limits_{n\rightarrow\infty} \frac{n}{\log n}(n^{\frac{1}{n}}-1)$? Here is  my attempt.
Put $f(x)=\frac{x}{\log x}(x^{\frac{1}{x}}-1)$ for $x>1$. Then
\begin{aligned}
\lim\limits_{n\rightarrow\infty} \frac{n}{\log n}(n^{\frac{1}{n}}-1) &= \lim\limits_{x\rightarrow\infty} f(x) \\
&= \lim\limits_{x\rightarrow\infty}\frac{x^{\frac{1}{x}}-1}{\frac{\log x}{x}} \\
&= \lim\limits_{x\rightarrow\infty} \frac{x^{\frac{1}{x}-2}(1-\log x)}{\frac{x^2+\log x}{x^4}} \\
&= \lim x^{\frac{1}{x}}\lim x^2\lim\frac{1-\log x}{x^2+\log x} \\
&= 1 \cdot (+\infty)\cdot (0)\mbox{.}
\end{aligned}
This method does not work.

As an alternative
$$\frac{n}{\log n}(n^{\frac{1}{n}}-1)=\frac{n^{\frac{1}{n}}-1}{\log \left(n^{\frac{1}{n}}\right)}$$
with $x+1=n^{\frac{1}{n}} \to 1$ that is as $x \to 0$
$$\lim_{x\to 0}\frac{x}{\log (1+x)}=1$$
