How do I find $\lim_{x \to 8} \frac{(\sqrt[3]{x} -2)}{x-8}$ by using the conjugate rule? I need to find: $\lim_{x \to 8} \frac{(\sqrt[3]{x} -2)}{x-8}$
I cannot solve this by substitution because that would cause the denominator to equal 0. 
Normally, I would simply use the conjugate trick, however I am uncertain how I would rationalize the numerator.  
$$\frac{(\sqrt[3]{x} -2)}{x-8}\times\frac{\sqrt[3]{x}+2}{\sqrt[3]{x}+2}$$
However, clearly this won't help me with anything, as I won't be able to factor anything. 
$$\frac{(\sqrt[3]{x^2} -4)}{(x-8)(\sqrt[3]{x^2}+2)}$$
I am unsure about how to continue from here. Perhaps I am on the wrong track entirely. Any form of guidance would be welcome. Thank you.  

You may apply L-Hospital Rule
$$L=\lim_{x \rightarrow 8} \frac{x^{1/3}-2}{x-8}= \lim_{x \rightarrow 8}(1/3)(x)^{-2/3}=\frac{1`}{12}.$$
