Find the
$$\lim_{n\to\infty}n(\sqrt[n]{n}-1)^2$$
maybe can use $$\sqrt[n]{n}=e^{\frac{1}{n}\ln{n}}=1+\dfrac{1}{n}\ln{n}+o(\dfrac{\ln{n}}{n})$$ so $$(\sqrt[n]{n}-1)^2\approx \dfrac{\ln^2{n}}{n^2}$$ so $$\lim_{n\to\infty}n(\sqrt[n]{n}-1)^2=\lim_{n\to\infty}\dfrac{\ln^2{n}}{n}=0?$$
My methods is true? and have other methods? Thank you
$$\begin{align} \ln \left(\lim_{n\to\infty}n(\sqrt[n]{n}-1)^2\right) &= \lim_{n\to\infty}\left(\ln n + 2 \ln\left(\sqrt[n]{n}-1\right)\right) \\ &= \lim_{n\to\infty}\left(\ln n + 2 \ln \left(e^\frac{\ln n}{n} - 1\right)\right) \\ &= \lim_{n\to\infty}\left(\ln n + 2 \ln\left(1 + \dfrac{\ln n}{n} + O\left(\dfrac{\ln^2 n}{2n^2}\right)-1\right)\right) \\ &= \lim_{n\to\infty}\left(\ln n + 2 \ln\left(\dfrac{\ln n}{n}\right)\right) \\ &= \lim_{n\to\infty} \left(\ln n + 2 \ln \ln n - 2 \ln n\right) \end{align}$$
The logarithm diverges towards negative infinity (since $\ln n \gg \ln \ln n$). So the limit approaches zero.
