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

There are some good answers here, but let me share a method that is more computational (in that it doesn't require noticing certain inequalities and using the squeeze theorem), and may help you with other similar limits.
First, one convenient way of taking exponents out of a limit expression is to take a logarithm:
$$\lim_{n \rightarrow \infty} f(n)^{g(n)} = \exp\left[\lim_{n \rightarrow \infty} g(n)\ln(f(n))\right]$$
Provided that $f(n) > 0$ (and either limit exists). In your case, this becomes:
$$\lim_{n \rightarrow \infty} (1 + \frac{1}{n^2})^{n} = \exp\left[\lim_{n \rightarrow \infty} n\ln(1 + \frac{1}{n^2})\right]$$ 
To evaluate the limit inside the brackets on the right, you can use l'Hospital's rule. You need to change the indeterminate form to $0/0$ and replace $n$ with a 'continuous variable' $x$:
$$\lim_{n \rightarrow \infty} n \ln(1 + \frac{1}{n^2}) = \lim_{x \rightarrow \infty} \frac{\ln(1 + \frac{1}{x^2})}{\frac{1}{x}} = \lim_{x \rightarrow \infty} \frac{(-\frac{2}{x^3})/(1 + \frac{1}{x^2})}{-\frac{1}{x^2}} = \lim_{x \rightarrow \infty} \frac{-\frac{2}{x^3}}{-\frac{1}{x^2}(1 + \frac{1}{x^2})}$$
To simplify this last fraction, multiply the numerator and denominator by $-x^4$:
$$= \lim_{x \rightarrow \infty} \frac{2x}{x^2 + 1} = 0$$
Finally, your answer is:
$$\exp\left[\lim_{n \rightarrow \infty} n\ln(1 + \frac{1}{n^2})\right] = \exp[0] = 1$$
