Limit of $\frac{(n+1)^{2n}}{(n^2+1)^n}$ as $n\to \infty$

Question

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.

Answer 1

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 $$ \left(1+\frac{2n}{n^2+1}\right)^n \ge \left(1+\frac{2}{n+1}\right)^n = \frac{\left(1+\frac{2}{n+1}\right)^{n+1}}{\left(1+\frac{2}{n+1}\right)^{\hphantom{n+1}}} \to \frac{e^2}{1} = e^2 $$

Answer 2

$$\frac{\lim_{n\to\infty}(1+\frac1n)^{2n}}{\lim_{n\to\infty}(1+\frac1{n^2})^n}$$ Can you see the numerator is $e^2$ and the denominator is $1$?

Answer 3

Notice, $$\lim_{n\to \infty}\frac{(n+1)^{2n}}{(n^2+1)^n}=\lim_{n\to \infty}\frac{n^{2n}\left(1+\frac{1}{n}\right)^{2n}}{n^{2n}\left(1+\frac{1}{n^2}\right)^n}$$ $$=\lim_{n\to \infty}\frac{\left(1+\frac{1}{n}\right)^{2n}}{\left(1+\frac{1}{n^2}\right)^n}$$ $$=\frac{\lim_{n\to \infty}\left(1+\frac{1}{n}\right)^{2n}}{\lim_{n\to \infty}\left(1+\frac{1}{n^2}\right)^n}$$ $$=\frac{\left(\lim_{n\to \infty}\left(1+\frac{1}{n}\right)^{n}\right)^2}{\left(\lim_{n\to \infty}\left(1+\frac{1}{n^2}\right)^{n^2}\right)^{1/n}}$$ $$=\frac{\left(e\right)^2}{(e)^{0}}=\color{red}{e^{2}}$$

Answer 4

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.