$\displaystyle\lim_{n\to \infty} n^2(\sqrt[n]{2}-\sqrt[n+1]{2})$ i cannot figure out a way to find this limit. $$\displaystyle\lim_{n\to \infty} n^2(\sqrt[n]{2}-\sqrt[n+1]{2})$$
Its undeterminate form $0\cdot\infty$ so i tried using $$\displaystyle\lim_{n\to \infty}\frac{a^{X_n}-1}{X_n}=\ln{a}$$ this leads to $0\cdot\infty$ again. 
I then tried to transform it in $\frac{0}{0}$ $$\displaystyle\lim_{n\to \infty} \frac{\sqrt[n]{2}-\sqrt[n+1]{2}}{\frac{1}{n^2}}$$ here i dont know what i could do to find the limit.

We have: 
$
\begin{align*}
\lim_{n\to \infty} n^2(\sqrt[n]{2}-\sqrt[n+1]{2}) & = \lim_{n\to \infty} n^2\left( 2^{\frac{1}{n}} - 2^{\frac{1}{n+1}} \right) = \lim_{n\to \infty} n^2 2^{\frac{1}{n+1}} \left( 2^{\frac{1}{n}-\frac{1}{n+1}} - 1 \right) \\
& = \lim_{n\to \infty} n^2 2^{\frac{1}{n+1}} \left( 2^{\frac{1}{n^2+n}} - 1 \right) \\
& = \lim_{n\to \infty} \cfrac{n^2 2^{\frac{1}{n+1}} \left( 2^{\frac{1}{n^2+n}} - 1 \right)}{\cfrac{1}{n^2+n}(n^2+n)} \\
& = \lim_{n\to \infty} \dfrac{n^2}{n^2+n} \cdot \lim_{n\to \infty} 2^{\frac{1}{n+1}} \cdot 
\lim_{n\to \infty} \cfrac{2^{\frac{1}{n^2+n}} - 1}{\cfrac{1}{n^2+n}} \\
& = 1 \cdot 1 \cdot \ln 2 \\
& = \boxed{\ln 2}.
\end{align*}
$
