What does this function show at x = 5? f(x)=x^2−25/x−5

Question

A) removable discontinuity
B) jump discontinuity
C) infinite discontinuity
D) continuity
E) none of the above

Answer 1

at x=5, f(x) = 0/0, or undefined.

At any other value, though,
f(x) = (x-5)(x+5)/(x-5) = x+5

So, f(x) is undefined only at x=5. By defining f(5) = 10, the discontinuity is removed. f(x) = x+5 for all x.

Answer 2

So would it be none of the above?

Answer 3

Ummm. Do you not think it would be A?
I explained how the discontinuity could be removed. Hence, it is removable.