AP Calculus BC · Unit 1: Limits & Continuity
1.13 Removing Discontinuities
What is a removable discontinuity?
A removable discontinuity occurs at a point where the limit of a function exists but either the function is undefined there or its value doesn't match the limit. Graphically, this appears as a hole in the graph. It's called "removable" because you can fix it by redefining the function's value at that single point.
Finding the value that removes the hole
To remove a discontinuity at x = a, find lim(x→a) f(x). If this limit exists (both one-sided limits are equal and finite), define a new function g(x) equal to f(x) everywhere except at x = a, where g(a) = the limit value. This makes g continuous at a.
Common algebraic technique
Removable discontinuities often arise from a common factor in the numerator and denominator of a rational function that cancels out, like (x²-4)/(x-2) at x=2. Factor, simplify, then evaluate the limit by substitution. This reveals the value the function "should" have had, which becomes the redefined point.
Key terms
- Removable discontinuity
- A point where the limit exists but the function is undefined or doesn't equal the limit; appears as a hole in the graph.
- How to remove a discontinuity
- Redefine the function's value at that point to equal the limit, making it continuous there.
- Common cause of removable discontinuities
- A factor that cancels between the numerator and denominator of a rational function.
- Condition for removability
- The two-sided limit at that point must exist and be finite (one-sided limits are equal).
- Example: f(x) = (x²-4)/(x-2)
- Simplifies to x+2 for x≠2; limit as x→2 is 4, so defining f(2)=4 removes the discontinuity.
Try an AP-style question
f(x) = (x² + x − 12)/(x − 3). What value of f(3) removes the discontinuity?
- 7
- 3
- −4
- 0
Show the answer
A. x² + x − 12 = (x + 4)(x − 3), so the limit is 3 + 4 = 7.
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