# How do you find the limit lim_(x->3^+)|3-x|/(x^2-2x-3) ?

Oct 5, 2014

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

Note that since $x$ approaches $3$ from the right, $\left(3 - x\right)$ is negative, which means that $| 3 - x | = - \left(3 - x\right) = x - 3$.

by removing the absolute value sign and factoring out the denominator,

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

by cancelling out $\left(x - 3\right)$'s,

$= {\lim}_{x \to {3}^{+}} \frac{1}{x + 1} = \frac{1}{3 + 1} = \frac{1}{4}$

I hope that this was helpful.