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.
 A: 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$
A: 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 $b^2 = 2^{2/3}= 4^{1/3}$
A: Hint:
It's $\lim_\limits{y\to\sqrt[3]2}\dfrac{y-\sqrt[3]2}{y^9-8}=\lim_\limits{y\to\sqrt[3]2}\dfrac{y-\sqrt[3]2}{(y^3-2)(y^6+2y^3+4)}$
$$=\lim_\limits{y\to\sqrt[3]2}\dfrac{y-\sqrt[3]2}{(y-\sqrt[3]2)(y^2+\sqrt[3]2y+\sqrt[3]2^{2})(y^6+2y^3+4)}$$
A: $\textbf{Hint:}$
If you multiply the top and bottom by 
$$\sqrt[3]{x^2}+\sqrt[3]{2x}+\sqrt[3]{2^2}$$ 
the numerator turns into $x-2$.
A: $$\lim_{x\to 2}\frac{\sqrt[3]{x}-\sqrt[3]{2}}{x^3-8}=$$
$$\lim_{x\to 2}\frac{\sqrt[3]{x}-\sqrt[3]{2}}{(x-2)(x^2+2x+4)}= $$
$$\lim_{x\to 2}\frac{(\sqrt[3]{x}-\sqrt[3]{2})}{(\sqrt[3]{x}-\sqrt[3]{2})(\sqrt[3]{x^2}+\sqrt[3]{4} +\sqrt[3]{x}\sqrt[3]{2})(x^2+2x+4)}= $$
$$\lim_{x\to 2}\frac{1}{(\sqrt[3]{x^2}+\sqrt[3]{4} +\sqrt[3]{x}\sqrt[3]{2})(x^2+2x+4)}=\frac {1}{36\sqrt[3] 4} $$
A: 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!!!  
A: 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
$$\sqrt[3]{t+2}=\sqrt[3]{2}+\frac{t}{3\ 2^{2/3}}-\frac{t^2}{18\ 2^{2/3}}+O\left(t^3\right)$$
$$\frac{\sqrt[3]{t+2}-\sqrt[3]{2}}{t \left(t^2+6 t+12\right)}=\frac{\frac{t}{3\ 2^{2/3}}-\frac{t^2}{18\ 2^{2/3}}+O\left(t^3\right) }{t \left(t^2+6 t+12\right)}=\frac{\frac{1}{3\ 2^{2/3}}-\frac{t}{18\ 2^{2/3}}+O\left(t^2\right) }{12+6 t+t^2}\to \frac{1}{36\ 2^{2/3}}$$
