A limit about $\left(1+\frac{1}{n}\right)^{n}$? Here is my question:
$$\displaystyle\lim_{n\rightarrow \infty}n^2\left[\left(1+\frac{1}{1+n}\right)^{n+1}-\left(1+\frac{1}{n}\right)^{n}\right]=?$$
Any hints will be fine. Thank you!
 A: HINT
Note that (to be made more rigorous)


*

*$(1+\frac{1}{1+n}) ^{n+1}=e^{(n+1)\log\left(1+\frac{1}{1+n}\right)}\sim e^{(n+1)\left(\frac{1}{1+n}-\frac{1}{2(1+n)^2}\right) }=e^{1-\frac{1}{2(1+n)}}\sim e\left(1-\frac{1}{2(1+n)}\right)$

*$(1+\frac{1}{n}) ^{n}=e^{n\log\left(1+\frac{1}{n}\right)}\sim e^{n\left(\frac{1}{n}-\frac{1}{2n^2}\right) }=e^{1-\frac{1}{2n}}\sim e\left(1-\frac{1}{2n}\right)$
then
$$n^2\left[\left(1+\frac{1}{1+n}\right)^{n+1} - \left(1+\frac{1}{n}\right)^{n}\right]\sim e\cdot n^2\left(\frac{1}{2n}-\frac{1}{2(1+n)}\right)=e\cdot n^2\left(\frac{2}{4n^2+4n}\right)$$
A: Here is "any" hint:
Set $x=1/n$ and you will have
$$
\lim_{x\to 0^+}\frac{1}{x^2}\biggl(\Bigl(1+\frac{x}{1+x}\Bigr)^{1+1/x}-(1+x)^{1/x}\biggr).
$$
The expression
$$
\Bigl(1+\frac{x}{1+x}\Bigr)^{1+1/x}-(1+x)^{1/x}
$$
in the right-hand side can be expanded around $x=0$ (perhaps after defining it as $0$ at $x=0$).
A: Hint: Use 
$$n^2\left[\left(1+\frac{1}{1+n}\right)^{n+1}-\left(1+\frac{1}{n}\right)^{n}\right]=\frac{\left(1+\frac{1}{1+n}\right)^{n+1}-\left(1+\frac{1}{n}\right)^{n}}{\frac1{n^2}}$$
and then use L'Hopital's Rule.
