<s>[INST] Find $\lim_{n\rightarrow\infty} \frac{(n+1)^{2n^2+2n+1}}{(n+2)^{n^2+2n+1} n^{n^2}}$ Woflram gives $\frac{1}{e}$ as the limit, but I failed to obtain it. Please help.
$$\lim_{n\rightarrow\infty} \frac{(n+1)^{2n^2+2n+1}}{(n+2)^{n^2+2n+1}n^{n^2}}$$
 [/INST] \begin{align}
\lim_{n\rightarrow\infty} \frac{(n+1)^{2n^2+2n+1}}{(n+2)^{n^2+2n+1}n^{n^2}}
&=\lim_{n\rightarrow\infty} \frac{(n+1)^{2n^2}(n+1)^{2n+1}}{n^{n^2}(n+2)^{n^2}(n+2)^{2n+1}}\\
&=\lim_{n\rightarrow\infty} \frac{(n+1)^{2n^2}(n+1)^{2n}(n+1)}{\left[n(n+2)\right]^{n^2}(n+2)^{2n}(n+2)}\\
&=\lim_{n\rightarrow\infty} \left(\frac{(n+1)^2}{n(n+2)}\right)^{n^2}\cdot \lim_{n\rightarrow\infty} \left(\frac{n+1}{n+2}\right)^{2n}\cdot\lim_{n\rightarrow\infty}\frac{n+1}{n+2}\\
&=\lim_{n\rightarrow\infty} \left(\frac{n^2+2n+1}{n^2+2n}\right)^{n^2}\cdot
\left[\lim_{n\rightarrow\infty} \left(\frac{n+1}{n+2}\cdot\frac{\frac{1}{n}}{\frac{1}{n}}\right)^{n}\right]^2\cdot\lim_{n\rightarrow\infty}\left(\frac{n+1}{n+2}\cdot\frac{\frac{1}{n}}{\frac{1}{n}}\right)\\
&=\lim_{n\rightarrow\infty} \left(1+\frac{1}{n^2+2n}\right)^{n^2}\cdot
\left[\lim_{n\rightarrow\infty} \left(\frac{1+\frac{1}{n}}{1+\frac{2}{n}}\right)^{n}\right]^2\cdot\lim_{n\rightarrow\infty}\frac{1+\frac{1}{n}}{1+\frac{2}{n}}\\
&=\lim_{n\rightarrow\infty} \left(1+\frac{1}{n^2+2n}\right)^{n^2}\cdot
\left[\frac{\lim_\limits{n\rightarrow\infty} \left(1+\frac{1}{n}\right)^{n}}{\lim_\limits{n\rightarrow\infty} \left(1+\frac{2}{n}\right)^{n}}\right]^2\cdot\lim_{n\rightarrow\infty}\frac{1+0}{1+0}\\
&=\lim_{n\rightarrow\infty} \left(1+\frac{1}{n^2+2n}\right)^{n^2}\cdot
\left(\frac{e}{e^2}\right)^2\cdot 1
\end{align}
Now, as $n$ approaches infinity, the ratio of the quantities $n^2+2n$ and $n^2$ approaches $1$. What this means for us is that somewhere out there in infinity they want to become the same quantity and we can treat them as such:
$$
\lim_{x\rightarrow\infty} \left(1+\frac{1}{x}\right)^{x}\cdot\frac{1}{e^2}=
e\cdot\frac{1}{e^2}=\frac{1}{e}
$$
</s>