<s>[INST] Prove that $\lim\limits_{n\to \infty}\frac{n^2+n-1}{3n^2+1}=\frac{1}{3}$ using the definition 
Prove $$\lim_{n\to \infty}\frac{n^2+n-1}{3n^2+1}=\frac{1}{3}$$ using the definition.

Let there be $\varepsilon>0$ we need to find $N<n$ such that $\Big|\frac{n^2+n-1}{3n^2+1}-\frac{1}{3}$$
\Big|<\varepsilon$
$$\left|\frac{n^2+n-1}{3n^2+1}-\frac{1}{3}
\right|=\left|\frac{3n^2+3n-3-3n^2-1}{9n^2+3}
\right|=\left|\frac{3n-4}{9n^2+3}\right|$$
What should I do next, can I say that because $n\to \infty$ we can neglect $-4$ and $3$ from the expression?
 [/INST] Hint: $\frac{n-3}{3n^2+1}\leq \frac{1}{3n}$
</s>