# How do you find the limit of (x-5)/(x^2-25) as x->5?

$\frac{1}{10}$
$\frac{x - 5}{{x}^{2} - 5} = \frac{x - 5}{\left(x - 5\right) \left(x + 5\right)} = \frac{1}{x + 5} , \text{ for each } x \ne 5$
${\lim}_{x \to 5} f \left(x\right) = {\lim}_{x \to 5} \frac{1}{x + 5} = \frac{1}{10}$