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.

Hint: Consider $f(x)=\sqrt[3]{x}$.  So $$f'(2)=\lim_{x \to 2}\frac{\sqrt[3]{x}-\sqrt[3]{2}}{x-2}=\frac{1}{3}x^{-\frac{2}{3}}\Bigg|_{x=2}=\frac{1}{3}\frac{1}{2^{\frac{2}{3}}}$$ So $$\frac{\sqrt[3]{x}-\sqrt[3]{2}}{x^3-8}=\frac{\sqrt[3]{x}-\sqrt[3]{2}}{x-2} \frac{1}{x^2+2x+4} \longrightarrow f'(2).\frac{1}{12}$$ as $ x \to 2$
