Proving limits without limit theorems I have to prove the following limit without using any limit theorems. I can only do so by using the Archimedean Proprety and the definition of a limit.
I have to prove:
$$\lim_{n\to\infty} \frac{n^2 - 2}{3n^2 + 1} = \frac13$$

Okay, so lets look at the problem:
$$\lim_{n\to \infty} \frac{n^2 -2}{3n^2 +1}.$$
The key here is to do a bit of algebra - on that basis, remove a factor of $n^2$ on the top and bottom:
$$\lim_{n\to \infty} \frac{n^2\big(1-\frac{2}{n^2}\big)}{n^2\big(3 +\frac{1}{n^2}\big)} = \lim_{n\to \infty} \frac{\big(1-\frac{2}{n^2}\big)}{\big(3 +\frac{1}{n^2}\big)}.$$
Now we can apply the limit laws step by step (I don't know how familiar you are with this, so I'll go slowly), the first is that we can pass the limit over the fraction:
$$\lim_{n\to \infty} \frac{\big(1-\frac{2}{n^2}\big)}{\big(3 +\frac{1}{n^2}\big)} = \frac{\lim_{n\to \infty} \big(1-\frac{2}{n^2}\big)}{\lim_{n\to \infty} \big(3 +\frac{1}{n^2}\big)}.$$
Then we can pass the limit through the brackets (baring in mind that the limit of a constant is just the constant):
$$\frac{\big(1-\lim_{n\to \infty} \big[ \frac{2}{n^2}\big]\big)}{ \big(3 +\lim_{n\to \infty}\big[ \frac{1}{n^2}\big]\big)}.$$
Knowing now that $\lim_{n\to \infty} \frac{1}{n^2} = 0$ we have that
$$\lim_{n\to \infty} \frac{n^2 -2}{3n^2 +1} = \frac{1-0}{3+0} = \frac{1}{3}.$$
