<s>[INST] 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.
 [/INST] $$\lim_{x\to 2}\frac{\sqrt[3]{x}-\sqrt[3]{2}}{x^3-8}=$$
$$\lim_{x\to 2}\frac{\sqrt[3]{x}-\sqrt[3]{2}}{(x-2)(x^2+2x+4)}= $$
$$\lim_{x\to 2}\frac{(\sqrt[3]{x}-\sqrt[3]{2})}{(\sqrt[3]{x}-\sqrt[3]{2})(\sqrt[3]{x^2}+\sqrt[3]{4} +\sqrt[3]{x}\sqrt[3]{2})(x^2+2x+4)}= $$
$$\lim_{x\to 2}\frac{1}{(\sqrt[3]{x^2}+\sqrt[3]{4} +\sqrt[3]{x}\sqrt[3]{2})(x^2+2x+4)}=\frac {1}{36\sqrt[3] 4} $$
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