Computing $\lim\limits_{n\to\infty}(1+1/n^2)^n$

Why is $\lim\limits_{n\to\infty}(1+\frac{1}{n^2})^n = 1$?

Could someone elaborate on this? I know that $\lim\limits_{n\to\infty}(1+\frac{1}{n})^n = e$.

I know that $\lim_{n\to\infty}(1+\frac{1}{n})^n = e$

Indeed. A consequence of this statement that you know, is that there exists some finite $K\gt1$ such that, for every positive $n$, $$ 1\lt\left(1+\frac1n\right)^n\lt K, $$ (It happens that the optimal upper bound is $K=\mathrm e$, but, to solve your problem, one can forget such a refinement.) Hence, $$ 1\lt\left(1+\frac1{n^2}\right)^n=\left(\left(1+\frac1{n^2}\right)^{n^2}\right)^{1/n}\lt K^{1/n}. $$ Since $K^{1/n}\to1$ irrespectively of the value of $K$, this proves that $$ \lim_{n\to\infty}\left(1+\frac1{n^2}\right)^n=1. $$

Hint: prove that $$ \log (1+u) \le u $$

then use the continuity of $\exp$.

details:

$$\left(1+\frac 1{n^2} \right)^n =\exp \left(n \log\left(1+\frac 1{n^2} \right)\right) \\ 0\le n \log\left(1+\frac 1{n^2}\right) \le \frac 1n \\ 1\le \left(1+\frac 1{n^2} \right)^n \le\exp\frac 1n\to 1 $$

As you know that $$\lim_{n\to \infty} (1+\frac{1}{n})^n=e,$$ you also have $$\lim_{n\to \infty} (1+\frac{1}{n^2})^{n^2}=e,$$ which yields $$\lim_{n\to \infty} (1+\frac{1}{n^2})^{n}=\lim_{n\to \infty} \left((1+\frac{1}{n^2})^{n^2}\right)^{1/n}=\lim_{n\to \infty} e^{1/n}=1.$$ EDIT: The last two equalities should be seen together. It works because the interior of the parenthesis converges towards $e$ and you take the $n$-th root of some number closer and closer to $e$. When $n$ goes to infinity, this is the same as the value of the limit of $e^{1/n}$.

It's really not necessary to know anything about the subtler limit, $e=\lim_{n\to\infty}\left(1+{1\over n}\right)^n$. A binomial expansion and some crude estimates are sufficient here. Note first that

$${n\choose k}\left({1\over n}\right)^k={n\over n}\cdot{n-1\over n}\cdot\cdots\cdot{1\over n}\le1$$

and therefore

$$\left(1+{1\over n^2}\right)^n=\sum_{k=0}^n{n\choose k}\left({1\over n^2}\right)^k=\sum_{k=0}^n{n\choose k}\left({1\over n}\right)^k{1\over n^k}\le\sum_{k=0}^n{1\over n^k}\le1+{1\over n}+{1\over n^2}(n-2)$$

The Squeeze Theorem takes over from here.