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 $\lim\limits_{n\to\infty}(1+\frac{1}{n})^n = e$.

You can do this without knowing anything about $e$. Using Bernoulli's Inequality we have $$\left(1 - \frac{1}{n^{4}}\right)^{n} \geq 1 - \frac{1}{n^{3}}$$ and clearly we have $$\left(1 - \frac{1}{n^{4}}\right)^{n} \leq 1 - \frac{1}{n^{4}}$$ so that $$1 - \frac{1}{n^{3}} \leq \left(1 - \frac{1}{n^{4}}\right)^{n} \leq 1 - \frac{1}{n^{4}}\tag{1}$$ and via Squeeze theorem we get $$\lim_{n \to \infty}\left(1 - \frac{1}{n^{4}}\right)^{n} = 1\tag{2}$$ Using similar technique we get the inequality $$1 - \frac{1}{n} \leq \left(1 - \frac{1}{n^{2}}\right)^{n} \leq 1 - \frac{1}{n^{2}}$$ and via Sqeeze theorem we get $$\lim_{n \to \infty}\left(1 - \frac{1}{n^{2}}\right)^{n} = 1\tag{3}$$ and dividing $(2)$ by $(3)$ we get $$\lim_{n \to \infty}\left(1 + \frac{1}{n^{2}}\right)^{n} = 1\tag{4}$$
