limit involving $e$, ending up without $e$. Compute the limit
$$ \lim_{n \rightarrow \infty} \sqrt n \cdot \left[\left(1+\dfrac 1 {n+1}\right)^{n+1}-\left(1+\dfrac 1 {n}\right)^{n}\right]$$
we have a bit complicated solution using Mean value theorem. Looking for others

Clearly if $a_{n} = \left(1 + \dfrac{1}{n}\right)^{n}$ then we have
\begin{align}
L &= \lim_{n \to \infty}\sqrt{n}\{a_{n + 1} - a_{n}\}\notag\\
&= \lim_{n \to \infty}\sqrt{n}\left[\exp\left\{(n + 1)\log\left(1 + \frac{1}{n + 1}\right)\right\} - \exp\left\{n\log\left(1 + \frac{1}{n}\right)\right\}\right]\notag\\
&= \lim_{n \to \infty}\sqrt{n}\left(1 + \frac{1}{n}\right)^{n}\left[\exp\left\{(n + 1)\log\left(1 + \frac{1}{n + 1}\right) - n\log\left(1 + \frac{1}{n}\right)\right\} - 1\right]\notag\\
&= e\lim_{n \to \infty}\sqrt{n}\cdot t\cdot\frac{e^{t} - 1}{t}\text{ (where }t = (n + 1)\log\left(1 + \frac{1}{n + 1}\right) - n\log\left(1 + \frac{1}{n}\right) \to 0)\notag\\
&= e\lim_{n \to \infty}\sqrt{n}\left\{(n + 1)\log\left(1 + \frac{1}{n + 1}\right) - n\log\left(1 + \frac{1}{n}\right)\right\}\notag\\
&= e\lim_{n \to \infty}\sqrt{n}\left\{\log\left(1 + \frac{1}{n + 1}\right) + n\log\left(\frac{n(n + 2)}{(n + 1)^{2}}\right)\right\}\notag\\
&= e\lim_{n \to \infty}\sqrt{n}\left\{\log\left(1 + \frac{1}{n + 1}\right) + n\log\left(1 - \frac{1}{(n + 1)^{2}}\right)\right\}\notag\\
&= e\lim_{n \to \infty}\frac{1}{\sqrt{n}}\left\{n\log\left(1 + \frac{1}{n + 1}\right) + n^{2}\log\left(1 - \frac{1}{(n + 1)^{2}}\right)\right\}\notag\\
&= e\lim_{n \to \infty}\frac{1}{\sqrt{n}}\left\{\frac{n}{n + 1}\cdot(n + 1)\log\left(1 + \frac{1}{n + 1}\right) + \frac{n^{2}}{(n + 1)^{2}}\cdot(n + 1)^{2}\log\left(1 - \frac{1}{(n + 1)^{2}}\right)\right\}\notag\\
&= e\cdot 0\{1\cdot 1 - 1\cdot 1\}\notag\\
&= 0\notag
\end{align}
This is done without Taylor/Mean Value Theorem (essentially without using derivatives). If $\sqrt{n}$ is replaced by $n^{2}$ (as mentioned in Lucian's comment to the original question) then we will need these higher level tools.
