# How do you find the limit of (x^2 - 64) / ((x^(1/3)) -2) as x approaches 8?

Sep 11, 2016

$192$

#### Explanation:

First, factor the numerator as a difference of squares, which takes the form ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$.

${\lim}_{x \rightarrow 8} \frac{{x}^{2} - {8}^{2}}{{x}^{\frac{1}{3}} - 2}$

$= {\lim}_{x \rightarrow 8} \frac{\left(x + 8\right) \left(x - 8\right)}{{x}^{\frac{1}{3}} - 2}$

Now, factor $x - 8$ as a difference of cubes, which takes the form ${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$.

$= {\lim}_{x \rightarrow 8} \frac{\left(x + 8\right) \left({\left({x}^{\frac{1}{3}}\right)}^{3} - {2}^{3}\right)}{{x}^{\frac{1}{3}} - 2}$

$= {\lim}_{x \rightarrow 8} \frac{\left(x + 8\right) \left({x}^{\frac{1}{3}} - 2\right) \left({x}^{\frac{2}{3}} + 2 {x}^{\frac{1}{3}} + 4\right)}{{x}^{\frac{1}{3}} - 2}$

The ${x}^{\frac{1}{3}} - 2$ terms will cancel:

$= {\lim}_{x \rightarrow 8} \left(\left(x + 8\right) \left({x}^{\frac{2}{3}} + 2 {x}^{\frac{1}{3}} + 4\right)\right)$

Plug in $8$:

$= \left(8 + 8\right) \left({8}^{\frac{2}{3}} + 2 \left({8}^{\frac{1}{3}}\right) + 4\right)$

$= 16 \left(4 + 2 \left(2\right) + 4\right)$

$= 16 \left(12\right)$

$= 192$