The question is show $$f(x)=\frac{1}{x^{3}+1}$$ is continuous at $x=2$.
I've got it to $$\frac{|x-2||-x^{2}-2x-4|}{|9(x^{3}+1)|}$$
But from here i don't understand how i pick my values for $\delta$ and how to use this value to come up with the final answer.
I believe $|x-2|=\delta<1$ so $x\in(1,3)$ Does this mean $$0 < \delta < min{[3,\frac{18 \epsilon}{5}]}$$ or not because i am not sure at all.
Any help would be appreciated many thanks
Hint: Suppose that we first of all specify that $\delta\lt 1$. Then $1\lt x\lt 3$, and therefore $|-x^2-2x-4|\lt 19$ and $9|x^3+1|\gt 9$.