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

##### 1 Answer
Sep 25, 2016

Always try substitution first. (it won't work for this one.) When finding a limit of a fraction and in doubt, rationalize either the numerator or denominator.

#### Explanation:

$\frac{\sqrt{{x}^{2} - 9}}{x - 3}$

If we rationalize the numerator, we'll be able to factor and reduce, so that looks reasonable.

$\frac{\sqrt{{x}^{2} - 9}}{x - 3} \cdot \frac{\sqrt{{x}^{2} - 9}}{\sqrt{{x}^{2} - 9}} = \frac{{x}^{2} - 9}{\left(x - 3\right) \sqrt{{x}^{2} - 9}}$

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

$= \frac{x + 3}{\sqrt{{x}^{2} - 9}}$

Now, as $x$ approaches $3$, it must be on the left. (I am assuming that we want to stay in the real numbers.)

As $x \rightarrow {3}^{-}$, the numerator is approaching $6$ and the denominator is a positive number approaching $0$.

As $x \rightarrow {3}^{-}$, the ratio is increasing without bound.

${\lim}_{x \rightarrow {3}^{-}} \frac{\sqrt{{x}^{2} - 9}}{x - 3} = {\lim}_{x \rightarrow {3}^{-}} \frac{x + 3}{\sqrt{{x}^{2} - 9}} = \infty$