Limit of $\lim_{n\to \infty}\sqrt{n}\left(\left(1+\frac{1}{n+1}\right)^{n+1}-\left(1+\frac{1}{n}\right)^{n}\right)=?$?

Question

limit of $$\lim_{n\to \infty}\sqrt{n}\left(\left(1+\frac{1}{n+1}\right)^{n+1}-\left(1+\frac{1}{n}\right)^{n}\right)=?$$

Answer 1

$$\begin{align*}0\leq\sqrt{n}\left[\left(1+\frac{1}{n+1}\right)^{n+1}-\left(1+\frac{1}{n}\right)^{n}\right]&\leq\sqrt{n}\left[\left(1+\frac{1}{n}\right)^{n+1}-\left(1+\frac{1}{n}\right)^{n}\right]\\&\;\;\;\;\;=\frac{1}{\sqrt{n}}\left(1+\frac{1}{n}\right)^{n}\\&\quad\;\leq \frac{e}{\sqrt{n}}\end{align*}$$

Hence:

$$0\leq \lim_{n\to\infty}\sqrt{n}\left[\left(1+\frac{1}{n+1}\right)^{n+1}-\left(1+\frac{1}{n}\right)^{n}\right]\leq \lim_{n\to\infty}\frac{e}{\sqrt{n}}=0$$

Answer 2

$$\lim_{n\to \infty}\sqrt{n}\left(\left(1+\frac{1}{n+1}\right)^{n+1}-\left(1+\frac{1}{n}\right)^{n}\right)=\lim_{n\to \infty}\sqrt{n}\left(1+\frac{1}{n}\right)^{n}\left(\frac{\left(1+\frac{1}{n+1}\right)^{n+1}}{\left(1+\frac{1}{n}\right)^{n}}-1\right)=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n}\lim_{n\to\infty}\sqrt{n}\left(\frac{\left(1+\frac{1}{n+1}\right)^{n+1}}{\left(1+\frac{1}{n}\right)^{n}}-1\right)=e\lim_{n\to\infty}\left(\frac{\sqrt n}{n+1}\right)=0$$