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
 A: Taylor expansion always works:
\begin{align}
& \left(1 + \frac{1}{n + 1}\right)^{n + 1} - \left(1 + \frac{1}{n}\right)^{1/n}\\
= & \exp\left[(n + 1)\log\left(1 + \frac{1}{n + 1}\right)\right] - \exp\left[n\log\left(1 + \frac{1}{n}\right)\right] \\
= & \exp\left[(n + 1)\left(\frac{1}{n + 1} - \frac{1}{2(n + 1)^2} + o\left(\frac{1}{(n  + 1)^2}\right)\right)\right] - \exp\left[n \left(\frac{1}{n} - \frac{1}{2n^2} + o\left(\frac{1}{n^2}\right)\right)\right] \\
= & \exp\left[1 - \frac{1}{2(n + 1)} + o\left(\frac{1}{n + 1}\right)\right]
-\exp\left[1 - \frac{1}{2n} + o\left(\frac{1}{n}\right)\right] \\
= & \left\{e + e\left[1 - \frac{1}{2(n + 1)} + o\left(\frac{1}{n + 1}\right) - 1\right] + O\left(\frac{1}{(n + 1)^2}\right)\right\} \\ 
- & \left\{e + e\left[1 - \frac{1}{2n} + o\left(\frac{1}{n}\right) - 1\right] + O\left(\frac{1}{n^2}\right)\right\} \\
= & o\left(\frac{1}{n}\right)
\end{align}
Therefore the original expression is $o\left(\frac{1}{\sqrt{n}}\right)$, hence the limit is $0$.
A: First we have the simple observation:
$$\left ( 1+\frac{1}{n+1} \right )^{n+1} = \left ( 1+\frac{1}{n+1} \right )^n \left ( 1+\frac{1}{n+1} \right ).$$
Now we consider the function $\left ( 1+x \right )^n$. Its derivative is $n(1+x)^{n-1}$. Therefore 
$$\left | \left ( 1+\frac{1}{n} \right )^n - \left ( 1+\frac{1}{n+1} \right )^n \right | \leq \frac{n(1+\xi)^{n-1}}{n(n+1)} = \frac{(1+\xi)^{n-1}}{n+1}$$
where $\xi \in \left ( \frac{1}{n+1},\frac{1}{n} \right )$. Since we know the numerator here is bounded, this difference is at most on the order of $1/n$. Multiplying by the additional factor of $\left ( 1+\frac{1}{n+1} \right )$ merely adds another term of order at most $1/n$. So the difference, rescaled by $\sqrt{n}$, is of order at most $n^{-1/2}$, and hence goes to zero.
This is probably very similar to your mean value theorem solution, seeing as this is also a mean value theorem solution. But maybe it is cleaner than yours.
A surprising fact that I don't see how to prove so easily is that in fact the difference in the original problem is on the order of $n^{-2}$, meaning that my estimate above is actually quite poor. Apparently this is because there is a "major cancellation": the additional factor of $1+\frac{1}{n+1}$ "just barely" makes the difference positive rather than negative, in such a way that its order of magnitude drops.
A: 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.
A: As Zhanxiong answered, Taylor expansion always works.
Consider, for large value of $x$, $$A(x)=\big(1+\frac 1x\big)^x$$ $$\log\big(A(x)\big)=x\log\big(1+\frac 1x\big)=x\big( \frac{1}{x}-\frac{1}{2 x^2}+\frac{1}{3 x^3}+\cdots\big)=1-\frac{1}{2 x}+\frac{1}{3 x^2}+\cdots$$ $$A(x)=e \, e^{-\frac{1}{2 x}+\frac{1}{3 x^2}}$$ So $$A(n+1)-A(n)\approx e\Big(e^{-\frac{1}{2 (n+1)}+\frac{1}{3 (n+1)^2}} -e^{-\frac{1}{2 n}+\frac{1}{3 n^2}}\Big)$$ Now, using , for small $y$, $e^y=1+y+\frac{y^2}2+\cdots$ $$A(n+1)-A(n)\approx e\Big(\frac{1}{2 n^2}-\frac{17}{12 n^3}+\cdots\Big)$$
