## Calculus 10th Edition

A removable discontinuity at $x=5$ and a non-removable discontinuity at $x=-5$.
$f(x)=\dfrac{(x-5)}{(x-5)(x+5)}=\dfrac{1}{x+5}\to x\ne5$ and $x\ne-5$. Hence, we have an non-removable discontinuity at $x=-5$ and a removable discontinuity at $x=5.$ A graph of $\frac{1}{x+5}$ will make the situation clear.