<s>[INST] Limit of $\frac{(n+1)^{2n}}{(n^2+1)^n}$ as $n\to \infty$ So it is given to find $$\lim_{n\to \infty}\dfrac{(n+1)^{2n}}{(n^2+1)^n}$$
So what I did is
$$\lim_{n\to \infty}\dfrac{(n+1)^{2n}}{(n^2+1)^n}=\lim_{n\to \infty}\dfrac{(n^2+2n+1)^{n}}{(n^2+1)^n}=\lim_{n\to \infty}\left(1+\frac{2n}{n^2+1}\right)^n$$
Now the rightmost form, is it $e^2$? I mean I am unable convince myself that it is some form of exponential function. Help me out. 
 [/INST] Your guess is right.
We have
$$
1+\frac{2n}{n^2+1} \le 1+\frac{2n}{n^2} = 1+\frac{2}{n}
$$
and so
$$
\left(1+\frac{2n}{n^2+1}\right)^n \le \left(1+\frac{2}{n}\right)^n \to e^2
$$
On the other hand,
$$
1+\frac{2n}{n^2+1} \ge 1+\frac{2n}{n^2+n} = 1+\frac{2}{n+1}
$$
and so
$$
\left(1+\frac{2n}{n^2+1}\right)^n
\ge \left(1+\frac{2}{n+1}\right)^n
= \frac{\left(1+\frac{2}{n+1}\right)^{n+1}}{\left(1+\frac{2}{n+1}\right)^{\hphantom{n+1}}}
\to \frac{e^2}{1} = e^2
$$
</s>