$\displaystyle\lim_{n\to \infty} n^2(\sqrt[n]{2}-\sqrt[n+1]{2})$ 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.
 A: 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*}
$
A: Write
$$2^{\frac{1}{n}} = e^{\frac{\log{2}}{n}}$$
so that 
$$\begin{align} \lim_{n \rightarrow \infty} n^2 \left (2^{\frac{1}{n}} - 2^{\frac{1}{n+1}} \right ) &= \lim_{n \rightarrow \infty} n^2 \left (e^{\frac{\log{2}}{n}} - e^{\frac{\log{2}}{n+1}} \right ) \\ &= \lim_{n \rightarrow \infty} n^2 \log{2} \left ( \frac{1}{n} - \frac{1}{n+1}\right ) \\ &= \lim_{n \rightarrow \infty} \frac{n \log{2}}{n+1} \\ &= \log{2} \\ \end{align} $$
The second step relies on the fact that
$$e^x = 1+ x + O(x^2)$$
as $x \rightarrow 0$.
A: By the mean value theorem 
$$\displaystyle\lim_{n\to \infty} n^2(\sqrt[n]{2}-\sqrt[n+1]{2})=\displaystyle\lim_{n\to \infty} \left(n^2 \times\frac{2^{1/c_{n}}\ln 2}{c_n^2}\right)=\ln2$$
where $n<c_n<n+1$
Chris.
