# How do you find the limit of (3x + 9)/(x^2 - 9) as x approaches -3?

Jul 8, 2018

$- \frac{1}{2}$

#### Explanation:

Given: $\lim x \to - 3 \text{ of } f \left(x\right) = \frac{3 x + 9}{{x}^{2} - 9}$

First factor both the numerator and denominator. Note that the denominator is the difference of squares, $\left({a}^{2} - {b}^{2}\right) = \left(a - b\right) \left(a + b\right)$

$f \left(x\right) = \frac{3 x + 9}{{x}^{2} - 9} = \frac{3 \cancel{\left(x + 3\right)}}{\cancel{\left(x + 3\right)} \left(x - 3\right)} = \frac{3}{x - 3}$

Now take the limit at $x = - 3$

$\lim x \to - 3 \text{ of } f \left(- 3\right) = \frac{3}{- 3 - 3} = \frac{3}{-} 6 = - \frac{1}{2}$

Jul 8, 2018

$- \frac{1}{2}$

#### Explanation:

${\lim}_{x \to - 3} \frac{3 x + 9}{{x}^{2} - 9}$

$= {\lim}_{x \to - 3} \frac{3 \cancel{\left(x + 3\right)}}{\cancel{\left(x + 3\right)} \left(x - 3\right)}$

$= {\lim}_{x \to - 3} \frac{3}{x - 3} = \frac{3}{- 6} = - \frac{1}{2}$