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.