Evaluating $\lim\limits_{x\to 2} \frac{\sqrt[3]{x} - \sqrt[3]{2}}{x^3 - 8}$ without l'Hospital's rule 
How to evaluate the following limit?
  $$
\lim\limits_{x\to 2} \dfrac{\sqrt[3]{x} - \sqrt[3]{2}}{x^3 - 8}
$$

I factored the denominator into $(x-2)(x^2+2x+4)$, but I couldn't go on from there.

would solve it like this:
Hint: You want to split limits into two, in order for the main limit to be defined, two sub-limits need to for a start be undefined, and that will be achieved like this:
$$\lim_{x\to 2} \frac{\frac{\sqrt[3]x-\sqrt[3]2}{x-2}}{\frac{x^3-8}{x-2}}=\frac{l1}{l2}$$
then I will solve just $l1$ but you will get an idea:
$$l1=\lim_{x\to 2}\frac{\sqrt[3]x-\sqrt[3]2}{x-2} \cdot \frac{\sqrt[3]{x^2+\sqrt[3]{2x}+\sqrt[3]4}}{\sqrt[3]{x^2+\sqrt[3]{2x}+\sqrt[3]4}}= \frac{x-2}{(x-2)(\sqrt[3]{x^2+\sqrt[3]{2x}+\sqrt[3]4})}=\frac{1}{3 \cdot \sqrt[3]4}$$
Now you do the same for the $l2$, divide it, and it's done. THIS IS MY FIRST ANSWER!!! GIVING BACK TO THE COMMUNITY!!!  
