Example 8: Consider the following function:
If you factor the top and bottom, you can see where the discontinuities are.
The function has a zero in the denominator when x = 1 or x = 5, so the function is discontinuous at those
two points. But you can cancel the term (x − 5) from both the numerator and the denominator, leaving you
with
Now the reduced function is continuous at x = 5. Thus, the original function has a removable discontinuity
at x = 5. Furthermore, if you now plug x = 5 into the reduced function, you get
The discontinuity is at x = 5, and there’s a hole at . In other words, if the original function were
continuous at x = 5, it would have the value . Notice that this is the same as f(x).
These are the types of discontinuities that you can expect to encounter on the AP Exam. Here are some
sample problems and their solutions. Cover the answers as you work, then check your results.
PROBLEM 1. Is the function f(x) = continuous at x = 2?
Answer: Test the conditions necessary for continuity.
Condition 1: f(2) = 9, so we’re okay so far.
Condition 2: The f(x) = 15 and the f(x) = 9. These two limits don’t agree, so the f(x)
doesn’t exist and the function is not continuous at x = 2.