Question

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

Answer 1

Apparently this function has two points of discontinuity for `x=+-4` where the denominator becomes zero giving a division by zero that cannot be performed.
One of these two values, however, can be eliminated considering that:
`(x-4)/(x^2-16)=cancel((x-4))/(cancel((x-4))(x+4))=1/(x+4)`
So that only `x=-4` maintains its character of discontinuity while `x=4` is a removable discontinuity.
Graphically:

graph{(x-4)/(x^2-16) [-16.02, 16.02, -8.01, 8.01]}