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

For $x > 3$, we can write
$| 3 - x \frac{|}{{x}^{2} - 2 x - 3} = \frac{x - 3}{\left(x - 3\right) \left(x + 1\right)} = \frac{1}{x + 1}$
lim_{x to 3^+}|3-x|/{x^2-2x-3} =lim_{x to 3^+}1/{x+1}=1/{3+1}=1/4