# How do you find the points of continuity for f(x)=(x-4)/(x^2-16)?

Apparently this function has two points of discontinuity for $x = \pm 4$ where the denominator becomes zero giving a division by zero that cannot be performed.
$\frac{x - 4}{{x}^{2} - 16} = \frac{\cancel{\left(x - 4\right)}}{\cancel{\left(x - 4\right)} \left(x + 4\right)} = \frac{1}{x + 4}$
So that only $x = - 4$ maintains its character of discontinuity while $x = 4$ is a removable discontinuity.