$\lim\limits_{n\rightarrow\infty} \frac{n}{\log n}(n^{\frac{1}{n}}-1)$ 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.
 A: Rewrite the limmand as
$$\frac{n}{\log n}\left(n^{\frac{1}{n}}-1\right) = \frac{n}{\log n}\left(e^{\frac{\log n}{n}}-1\right)$$
Then with $x = \frac{\log n}{n}$ we have
$$\lim_{x \to 0^+} \frac{e^x-1}{x} = 1 $$
the classic limit.
A: 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 $0$.
A: Let $n=e^x.$ Then  $n^{1/n}-1=$ $e^{x/e^x}-1=[x/e^x][1+(x/e^x)/2!+(x/e^x)^2/3!+(x/e^x)^3/4!+...]$
which lies between $x/e^x=(\log n)/n$ and $(x/e^x)(1+(x/e^x))=(\log n)/n+((\log n)/n)^2.$
Because if $x\ge 0$ then $0\le x/e^x<1$ so $$0\le  (x/e^x)/2!+(x/e^x)^2/3!+(x/e^x)^3/4!+...\le$$ $$\le [(x/e^x)/2!][1+1/2^1+1/2^2+...]=x/e^x.$$
A: 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 $x \to 0$
$$\lim_{x\to 0}\frac{x}{\log (1+x)}=1$$
