Finding the limit $\lim_{n\to\infty} \frac{n\left(\sqrt[n]{n}-1\right)}{\log n}$ I try to calculate the following limit:
$$\lim_{n\to\infty}\frac{n\left(\sqrt[n]{n}-1\right)}{\log n}$$
I think it should equal 1, because:
$$\exp(x)=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^{n}$$
(Already proven)
Solving for $x$ gives:
$$
\log x = \lim_{n \to \infty} n \left(\sqrt[n]{x}-1\right)
\implies \lim_{n\to\infty}\frac{n\left(\sqrt[n]{n}-1\right)}{\log n}=\lim_{n\to\infty}\frac{\log n}{\log n}=1
$$
But I like to calculate the limit with just standard things like L’Hôpital’s rule, because the previous way is maybe wrong and contains too much magic.
For example, I tried this:
$$
\begin{align*}
\lim_{n\to\infty}\frac{n\left(\sqrt[n]{n}-1\right)}{\log n}
&= \lim_{n\to\infty}\frac{\frac{d}{dn}\!\!\left(n\left(\sqrt[n]{n}-1\right)\right)}{\frac{d}{dn}\log n} \\[6pt]
&=\lim_{n\to\infty}\frac{n^{1/n-1}\left(-\left(\log\left(n\right)-1\right)\right)+n^{1/n}-1}{1/n} \\[6pt]
&=\lim_{n\to\infty}\left(n\left(n^{1/n-1}\left(-\left(\log\left(n\right)-1\right)\right)+n^{1/n}-1\right)\right) \\[6pt]
&=\lim_{n\to\infty}\left(n^{1/n}\left(-\left(\log\left(n\right)-1\right)\right)+n^{1/n+1}-n\right) \\[6pt]
&=???
\end{align*}
$$
But then it becomes really ugly and looks wrong. Have I done something wrong? Is there another way to find the limit?
Thank you for any ideas.
 A: Defining $\displaystyle a_n := \sqrt[n]{n} - 1$, we have $a_n > 1$ and the well-known result$\displaystyle \lim_{n \to \infty}a_n = \lim_{n \to \infty}\sqrt[n]{n} - 1 =0.$ 
Then
$$n = (1+a_n)^n, \\ \log n = n \log (1 + a_n).$$
Using the inequality $\displaystyle \frac{a_n}{1+a_n} \leqslant \log (1+a_n) \leqslant a_n$, we find
$$\frac{na_n}{1+a_n} \leqslant \log n  \leqslant na_n,\\ 1 \leqslant \frac{n a_n}{\log n}\leqslant 1 + a_n, \\ 1 \leqslant \frac{n (\sqrt[n]{n} -1)}{\log n}\leqslant \sqrt[n]{n} .$$
Tt follows from the squeeze principle that
$$\lim_{n \to \infty}\frac{n (\sqrt[n]{n} -1)}{\log n}=1.$$
Alternatively, using L'Hospital's rule for $x \in \mathbb{R}$,
$$\lim_{x\to\infty}\frac{x\left(\sqrt[x]{x}-1\right)}{\log x}=\lim_{x\to\infty}\frac{\exp{(x^{-1}\log x})-1}{x^{-1}\log x}\\=\lim_{x\to\infty}\frac{\exp{(x^{-1}\log x)})\frac{d}{dx}(x^{-1}\log x)}{\frac{d}{dx}(x^{-1} \log x )}\\=\lim_{x\to\infty}\exp(x^{-1} \log x)=1$$
Then it is not difficult to show, for $n \in \mathbb{N}$,
$$\lim_{n\to\infty}\frac{n\left(\sqrt[n]{n}-1\right)}{\log\left(n\right)}=\lim_{n\to\infty}\exp[\log(n)/n]=1.$$
A: $\begin{array}\\
\dfrac{n\left(\sqrt[n]{n}-1\right)}{\log n}
&=\dfrac{n\left(e^{\ln(n)/n}-1\right)}{\log n}\\
&=\dfrac{n\left(1+\frac{\ln(n)}{n}+O(\frac{\ln^2(n)}{n^2})-1\right)}{\log n}\\
&=\dfrac{n\left(\frac{\ln(n)}{n}+O(\frac{\ln^2(n)}{n^2})\right)}{\log n}\\
&=\dfrac{\left(\ln(n)+O(\frac{\ln^2(n)}{n})\right)}{\log n}\\
&=1+O(\frac{\ln(n)}{n})\\
&\to 1\\
\end{array}
$
A: rewrite your term in the form $\frac{n^{1/n}-1}{\log(n)}{n}$ and use L'Hospital the result is $1$
A: We can use approximations : $$\lim_{n\to\infty}\frac{n\left(\sqrt[n]{n}-1\right)}{\log n} = \lim_{n\to\infty}\frac{\left(\sqrt[n]{n}-1\right)}{\frac{\log n}{n}}=\lim_{n\to\infty}\frac{\left(n^{\frac{1}{n}}-1\right)}{\log n^{\frac{1}{n}}}$$
Now, observe that $\lim_{n\to\infty} \frac{\log n}{n} = 0$ and hence $\lim_{n\to\infty}n^{\frac{1}{n}}=1$. And we know $\log (x) \approx x-1$ when $x$ is close to $1$. Therefore $\log n^{\frac{1}{n}} \approx \left(n^{\frac{1}{n}}-1\right)$ and hence the required limit will be $1$
