# How do you evaluate the limit (x^2-9)/(x^3-27) as x approaches 3?

Sep 4, 2016

${\lim}_{x \rightarrow 3} \frac{{x}^{2} - 9}{{x}^{3} - 27} = \textcolor{g r e e n}{\frac{2}{9}}$

#### Explanation:

Two general equations to remember:
$\textcolor{w h i t e}{\text{XXX}} {a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$
$\textcolor{w h i t e}{\text{XXX}} {a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$

Therefore
$\textcolor{w h i t e}{\text{XXX}} \frac{{x}^{2} - 9}{{x}^{3} - 27} = \frac{\left(x - 3\right) \left(x + 3\right)}{\left(x - 3\right) \left({x}^{2} + 3 x + 9\right)}$

As long as $x \ne 3$ (which is true as long as $x$ is only approaching $3$)
the factors $\left(x - 3\right)$ can be cancelled out of the numerator an denominator.

So
$\textcolor{w h i t e}{\text{XXX}} {\lim}_{x \rightarrow 3} \frac{{x}^{2} - 9}{{x}^{3} - 27}$

$\textcolor{w h i t e}{\text{XXXXXX}} = {\lim}_{x \rightarrow 3} \frac{x + 3}{{x}^{2} + 3 x + 9}$

$\textcolor{w h i t e}{\text{XXXXXX}} = \frac{3 + 3}{9 + 9 + 9}$

$\textcolor{w h i t e}{\text{XXXXXX}} = \frac{6}{27}$

$\textcolor{w h i t e}{\text{XXXXXX}} = \frac{2}{9}$