# How do you find the limit of (x^2-9)/(x^2+2x-3) as x approaches 1^+?

Nov 8, 2016

${\lim}_{x \rightarrow {1}^{-}} \frac{{x}^{2} - 9}{{x}^{2} + 2 x - 3} = - \infty$

#### Explanation:

When we try to evaluate ${\lim}_{x \rightarrow {1}^{-}} \frac{{x}^{2} - 9}{{x}^{2} + 2 x - 3}$ by substitution we get a non-$0$ number over $0$. This tells us that we have some kind of infinite limits at $1$.

To determine whether the ratio is increasing or decreasing without bound (going to $\infty$ or $- \infty$) we need to investigate the signs of the factors.

${\lim}_{x \rightarrow {1}^{-}} \frac{{x}^{2} - 9}{{x}^{2} + 2 x - 3} = {\lim}_{x \rightarrow {1}^{-}} \frac{{x}^{2} - 9}{\left(x + 3\right) \left(x - 1\right)}$

As $x \rightarrow {1}^{+}$,

${x}^{2} - 9 \rightarrow - 8$, $\text{ }$ (The numerator is negative.)
$x + 3 \rightarrow 4$, $\text{ }$ (This factor is positive.)
$x - 1 \rightarrow {0}^{+}$, $\text{ }$ (This factor is positive.)

The ratio is decreasing without bound.

We write:

${\lim}_{x \rightarrow {1}^{-}} \frac{{x}^{2} - 9}{{x}^{2} + 2 x - 3} = - \infty$