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.

For a start, $$ x^9 - 1 = (x-1) (x^2+x+1)(x^6 + x^3 + 1)$$
This leads us to
$$ a^9 - b^9 = (a-b)(a^2 + ab + b^2)(a^6 + a^3 b^3 + b^6)  $$
Now take $a,b$ to be the two cube roots in your fraction
After you cancel the $(a-b)$ factors, you can simply set $a,b$ equal, so the denominator becomes $3 b^2 3 b^6 = 9 b^8$ with $b = 2^{1/3}.$ Now, to get all the way to Mohammad's answer, note $b^8 = b^6 b^2 = 4 b^2,$ while $b^2 = 2^{2/3}= 4^{1/3}$
