<s>[INST] $\lim\limits_{n\rightarrow\infty} \frac{n}{\log n}(n^{\frac{1}{n}}-1)$ 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.
 [/INST] The expression in your limit is $\frac{n^{1/n}-1}{\frac{\log n}{n}}$
Notice that  $\lim_{n\rightarrow\infty}\frac{\log n}{n}=0$,
and $n^{1/n}=\exp(\tfrac{1}{n}\log(n))$.
Setting $h_n=\frac{\log n}{n}$ yields
$$\frac{n^{1/n}-1}{(\log n)/n}=\frac{\exp(\tfrac{\log n}{n})-1}{\frac{\log n}{n}}=\frac{e^{h_n}-1}{h_n}\xrightarrow{n\rightarrow\infty}e^0=1$$
By definition of derivative of the exponential at $0$.
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