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.

Just another way.
$$\lim\limits_{x\to 2} \dfrac{\sqrt[3]{x} - \sqrt[3]{2}}{x^3 - 8}=\lim\limits_{t\to 0}\frac{\sqrt[3]{t+2}-\sqrt[3]{2}}{t \left(t^2+6 t+12\right)} $$ Now, using the binomial expansion or Taylor series
$$\sqrt[3]{t+2}=\sqrt[3]{2}+\frac{t}{3\ 2^{2/3}}-\frac{t^2}{18\ 2^{2/3}}+O\left(t^3\right)$$
$$\frac{\sqrt[3]{t+2}-\sqrt[3]{2}}{t \left(t^2+6 t+12\right)}=\frac{\frac{t}{3\ 2^{2/3}}-\frac{t^2}{18\ 2^{2/3}}+O\left(t^3\right) }{t \left(t^2+6 t+12\right)}=\frac{\frac{1}{3\ 2^{2/3}}-\frac{t}{18\ 2^{2/3}}+O\left(t^2\right) }{12+6 t+t^2}\to \frac{1}{36\ 2^{2/3}}$$
