> 
> Prove that $f$ is continuous at $x\_0$ using the $\delta$ and $\epsilon$ criteria for:
> 
> 
> $f(x)=\frac{x^3-2}{x+3}$, $x\_0=1$.
> 
> 
> 

I've made it halfway there, about; since $$|\frac{x^3-2}{x+3}+\frac{1}{4}|=|x-1||\frac{4x^2+4x+5}{4(x+3)}|$$

So then the issue is that we need $\delta>|x-1|$ but also bigger than the other expression. The only issue is, I don't really know how to find such a $\delta$. How can you proceed from here?