Evaluate $\lim\limits_{n\to\infty} \left(1+\frac{1}{n}\right)^{n^2}\cdot\left(1+\frac{1}{n+1}\right)^{-(n+1)^2}$ using the definition of $e$ 
Evaluate $\lim\limits_{n\to\infty} \left(1+\frac{1}{n}\right)^{n^2}\cdot\left(1+\frac{1}{n+1}\right)^{-(n+1)^2}.$

I know that $e^x=\lim\limits_{n\to\infty}\left(1+\dfrac{x}{n}\right)^n,$ so I need to somehow convert the limits to this form. I also noticed that $\left(1+\frac{1}{n+1}\right)=\left(\frac{n+2}{n+1}\right)=\left(\frac{n(n+2)}{(n+1)^2}\right)\cdot\left(1+\frac{1}{n}\right).$ 
Thus, the limit can be rewritten as $$\lim\limits_{n\to\infty}\dfrac{\left(1+\frac{1}{n}\right)^{n^2}}{\left(1+\frac{1}{n}\right)^{(n+1)^2}}\cdot\left(\dfrac{(n+1)^2}{n(n+2)}\right)^{(n+1)^2}\\
=\lim\limits_{n\to\infty}\left(1+\dfrac{1}{n}\right)^{-2n}\cdot\left(1+\dfrac{1}{n}\right)^{-1}\cdot\left(1+\dfrac{1}{n^2+2n}\right)^{n^2+2n}\cdot\left(1+\dfrac{1}{n^2+2n}\right)$$
$$=\dfrac{1}{e^2}\cdot(1)\cdot e\cdot(1)=\dfrac{1}{e}$$
 A: Your proof is nice, we have indeed
$$\left(1+\frac{1}{n}\right)^{n^2}\cdot\left(1+\frac{1}{n+1}\right)^{-(n+1)^2}=\\=\left(1+\frac{1}{n}\right)^{n^2}\cdot\left(1+\frac{1}{n+1}\right)^{-n^2}\cdot\left(1+\frac{1}{n+1}\right)^{-2n}\cdot\left(1+\frac{1}{n+1}\right)^{-1}=$$
$$=\left(\frac{(n+1)^2}{n(n+2)}\right)^{n^2}\cdot\left(1+\frac{1}{n+1}\right)^{-2n}\cdot\left(1+\frac{1}{n+1}\right)^{-1}=$$
$$=\left[\left(1+\frac{1}{n^2+2n}\right)^{n^2+2n}\right]^{\frac{n^2}{n^2+2n}}\cdot\left(1+\frac{1}{n+1}\right)^{-2n}\cdot\left(1+\frac{1}{n+1}\right)^{-1}\to e^1 \cdot e^{-2}\cdot 1 =\frac1e$$
A: Your work is correct. Just make sure you know how to prove that $e=\lim\limits_{n\to\infty}\left(1+\frac{1}{n}\right)^n.$ There's a nice geometric proof using areas under the curve.
A: This is not what I would consider an answer to the question but I decided to provide an in depth asymptotic analysis of the behaviour of this function in the limit as $x\to\infty$. Considering
$$f(x)=\left(1+\frac1x\right)^{x^2}\cdot\left(1+\frac1{x+1}\right)^{-(x+1)^2}$$
then we have
$$\begin{align}
\ln{(f(x))}
&=x^2\ln{\left(1+\frac1x\right)}-(x+1)^2\ln{\left(1+\frac1{x+1}\right)}\\
&=x^2\left(\frac1x-\frac1{2x^2}+\frac1{3x^3}-\frac1{4x^4}+o\left(\frac1{x^4}\right)\right)-(x+1)^2\left(\frac1{x+1}-\frac1{2(x+1)^2}+\frac1{3(x+1)^3}+\frac1{4(x+1)^4}+o\left(\frac1{x^4}\right)\right)\\
&=\left(x-\frac12+\frac1{3x}-\frac1{4x^2}+o\left(\frac1{x^2}\right)\right)-\left((x+1)-\frac1{2}+\frac1{3(x+1)}-\frac1{4(x+1)^2}+o\left(\frac1{x^2}\right)\right)\\
&=-1+\frac1{3x}-\frac1{4x^2}-\frac1{3(x+1)}+\frac1{4(x+1)^2}+o\left(\frac1{x^2}\right)\\
&=-1+\frac1{3x}\left(1-\frac1{1+\frac1x}\right)-\frac1{4x^2}\left(1-\frac1{(1+\frac1x)^2}\right)+o\left(\frac1{x^2}\right)\\
&=-1+\frac1{3x}\left(1-\left(1-\frac1x+\frac1{x^2}+o\left(\frac1{x^2}\right)\right)\right)-\frac1{4x^2}\left(1-\left(1-\frac2x+\frac3{x^2}+o\left(\frac1{x^2}\right)\right)\right)+o\left(\frac1{x^2}\right)\\
&=-1+\frac1{3x}\left(\frac1x-\frac1{x^2}+o\left(\frac1{x^2}\right)\right)-\frac1{4x^2}\left(\frac2x-\frac3{x^2}+o\left(\frac1{x^2}\right)\right)+o\left(\frac1{x^2}\right)\\
&=-1+\frac1{3x^2}+o\left(\frac1{x^2}\right)\\
\end{align}$$
$$\begin{align}
\therefore f(x)
&=\exp{\left(-1+\frac1{3x^2}+o\left(\frac1{x^2}\right)\right)}\\
&=\frac1e\cdot\exp{\left(\frac1{3x^2}+o\left(\frac1{x^2}\right)\right)}\\
&=\frac1e\left(1+\frac1{3x^2}+o\left(\frac1{x^2}\right)+o\left(\frac1{3x^2}+o\left(\frac1{x^2}\right)\right)\right)\\
&=\frac1e\left(1+\frac1{3x^2}+o\left(\frac1{x^2}\right)\right)\\
\end{align}$$
