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.  
 A: The denominator is already $0$ in the limit. That doesn't mean the limit doesn't exist.
Recall the factorization of a difference of two cubes:
$$
a^3 - b^3 = (a-b)(a^2+ab+b^2).
$$
In particular
$$
x-8 = \big(\sqrt[3] x - 2\big)\big( \sqrt[3]x^2 + 2\sqrt[3]x + 4\big).
$$
So
$$
\frac{\sqrt[3] x - 2}{x-8} = \frac{\sqrt[3]x-2}{\big(\sqrt[3]x-2\big)\big(\sqrt[3]x^2 +2\sqrt[3]x + 4\big)}.
$$
Can you take it from there?
A: Hint:
Using the formula for the difference of cubes,
$x-8=(\sqrt[3]x-2)(x^{2/3}+2\sqrt[3]{x}+4)$.
A: Take the steps below$$\lim_{x \to 8} \frac{\sqrt[3]{x} -2}{x-8}
=\lim_{x \to 8} \frac{(\sqrt[3]{x} -2)((\sqrt[3]{x})^2 +2\sqrt[3]{x} + 4)}{(x-8)((\sqrt[3]{x})^2 +2\sqrt[3]{x} + 4)}$$
$$=\lim_{x \to 8}\frac{x-8}{(x-8)((\sqrt[3]{x})^2 +2\sqrt[3]{x} + 4)}
=\lim_{x \to 8}\frac{1}{(\sqrt[3]{x})^2 +2\sqrt[3]{x} + 4}
=\frac1{4+4+4}=\frac1{12}$$
A: You may apply L-Hospital Rule
$$L=\lim_{x \rightarrow 8} \frac{x^{1/3}-2}{x-8}= \lim_{x \rightarrow 8}(1/3)(x)^{-2/3}=\frac{1`}{12}.$$
A: Let $f(x):=\sqrt[3]{x}.$ Then $ \frac{\sqrt[3]{x} -2}{x-8}= \frac{f(x)-f(8)}{x-8} \to f'(8)= \frac{1}{12}$ as $x \to 8.$
A: Just another way.
Let $x=y+8$ to make
$$A=\frac{\sqrt[3]{x} -2}{x-8}=\frac{\sqrt[3]{y+8} -2}{y}$$  Now, use the binomial expansion or Taylor series to get
$$A=\frac{1}{12}-\frac{y}{288}+O\left(y^2\right)$$ Back to $x$
$$A=\frac{1}{12}-\frac{x-8}{288}+O\left((x-8)^2\right)$$ WHich shows the limit and how it is approched.
Using your calculator, try for $x=9$, this would give as an exact result $A=0.0772$ while the above truncated formula gives $\frac{11}{144}=0.0764$
A: Set $y:=x^{1/3}$, and consider $y \rightarrow 2$.
We have $\dfrac{y-2}{y^3-8}$.
$y-2$ is a factor of $y^3 -8$, since $2^3-8=0$;
Exercise : polynomial long division.
$(y^3-8)÷(y-2)=y^2+2y+4$;
$-(y^3-2y^2)$

$2y^2-8$
$-(2y^2-4y)$

$4y-8$
$ -(4y-8)$

$0$;
Now consider $\lim_{y \rightarrow 2} \dfrac{1}{y^2+2y+4}$.
