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:
$$\lim_{n\to\infty} \frac{n^2 - 2}{3n^2 + 1} = \frac13$$

For $\epsilon >0$ sufficiently small $$\left|\dfrac{n^2-2}{3n^2+1}-\frac{1}{3}\right|<\epsilon \Leftrightarrow \left|\dfrac{-7}{9n^2+3}\right|<\epsilon $$ $$ \Leftrightarrow \dfrac{7}{9n^2+3}<\epsilon \Leftrightarrow \frac{7}{\epsilon}-3<9n^2\Leftrightarrow \frac{7-3\epsilon}{9\epsilon}<n^2\Leftrightarrow \sqrt{\frac{7-3\epsilon}{9\epsilon}}<n$$ That is, for every $\epsilon >0$ sufficiently small if we choose $$n_0=\left\lfloor\sqrt{\frac{7-3\epsilon}{9\epsilon}}\right\rfloor +1$$ then $$\left|\dfrac{n^2-2}{3n^2+1}-\frac{1}{3}\right|<\epsilon $$ if $n\ge n_0$ so, $$\lim_{n\to +\infty}\dfrac{n^2-2}{3n^2+1}=\dfrac{1}{3}.$$
