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. 
 A: $$\frac{\lim_{n\to\infty}(1+\frac1n)^{2n}}{\lim_{n\to\infty}(1+\frac1{n^2})^n}$$
Can you see the numerator is $e^2$ and the denominator is $1$?
A: Notice, $$\lim_{n\to \infty}\frac{(n+1)^{2n}}{(n^2+1)^n}=\lim_{n\to \infty}\frac{n^{2n}\left(1+\frac{1}{n}\right)^{2n}}{n^{2n}\left(1+\frac{1}{n^2}\right)^n}$$
$$=\lim_{n\to \infty}\frac{\left(1+\frac{1}{n}\right)^{2n}}{\left(1+\frac{1}{n^2}\right)^n}$$
$$=\frac{\lim_{n\to \infty}\left(1+\frac{1}{n}\right)^{2n}}{\lim_{n\to \infty}\left(1+\frac{1}{n^2}\right)^n}$$
$$=\frac{\left(\lim_{n\to \infty}\left(1+\frac{1}{n}\right)^{n}\right)^2}{\left(\lim_{n\to \infty}\left(1+\frac{1}{n^2}\right)^{n^2}\right)^{1/n}}$$
$$=\frac{\left(e\right)^2}{(e)^{0}}=\color{red}{e^{2}}$$
A: 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
$$
\left(1+\frac{2n}{n^2+1}\right)^n
\ge \left(1+\frac{2}{n+1}\right)^n
= \frac{\left(1+\frac{2}{n+1}\right)^{n+1}}{\left(1+\frac{2}{n+1}\right)^{\hphantom{n+1}}}
\to \frac{e^2}{1} = e^2
$$
A: Another way starting from what you wrote  $$A=\left(1+\frac{2n}{n^2+1}\right)^n$$ Take logarithms $$\log(A)=n \log\left(1+\frac{2n}{n^2+1}\right)$$ Now remember that, when $x$ is small $\log(1+x)=x+O\left(x^2\right)$. So, $$\log(A)\approx n \times\frac{2n}{n^2+1}=\frac{2n^2}{n^2+1}$$
I am sure that you can take from here.
