AP® Calculus AB review sheet from Aim for Five (aimforfive.com/calc-ab/units/1/1-13)
Unit 1 · Topic 1.13
1.13 Removing Discontinuities
If a function has a removable discontinuity, you can fix it by redefining the function at that one point to equal the limit. For a piecewise function, you can choose a constant that makes the pieces meet. Both are really the same idea: make the limit and the value match.
Key terms
- removable discontinuity
- redefining a function at a point
- piecewise function
- matching one-sided limits
Filling a hole
Suppose lim (x→c) f(x) = L exists but f(c) is missing or wrong. Define a new function that equals f(x) for x ≠ c and equals L at x = c. The new function is continuous at c.
You can only do this when the limit exists. A jump or an infinite discontinuity can't be removed by changing a single point.
A good way to see it: the original graph has a gap at one point. Since the curve on both sides heads into the same spot, there's exactly one height that patches it, which is the value of the limit. Any other value leaves a break.
A classic case: sin x / x is undefined at x = 0, but lim (x→0) sin x / x = 1, so defining the value at 0 to be 1 gives a function that is continuous everywhere.
Matching pieces with a constant
A piecewise function with an unknown constant, like k, is continuous at a break point c exactly when the left-hand limit equals the right-hand limit (and the function value, which comes from one of the pieces). Set the two pieces equal at x = c and solve for k.
You don't need a separate equation for f(c). The value at c comes from whichever piece includes c, so it automatically equals that piece's one-sided limit. Matching the two one-sided limits is enough.
If there are two unknowns, one continuity condition isn't enough. You'll need a second condition, often that the function is also differentiable there, so the slopes match too (2.4).
Why only removable breaks can be fixed
At a jump, the left side and right side head to different heights. Whatever single value you assign at c, at least one side won't connect to it. At an infinite discontinuity, the outputs don't approach any height at all, so there's nothing to fill in.
That's why the first step is always to find the limit. If it exists, the fix is to set f(c) equal to it. If it doesn't, no change to f(c) can make f continuous at c.
Writing the answer
When asked to define f(c) so that f is continuous, state the value and the reason: “Define f(3) = 7, because lim (x→3) f(x) = 7.” When asked for a constant, show the equation you set up, not just the number.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Redefine at one point
f(x) = (x² + x − 12) / (x − 3). What value should f(3) be given so that f is continuous at x = 3?
Show the solutionHide the solution
- Step 1: Substitute x = 3: (9 + 3 − 12) / 0 = 0/0, so look for a removable discontinuity.
- Step 2: Factor: x² + x − 12 = (x − 3)(x + 4). For x ≠ 3, f(x) = x + 4.
- Step 3: lim (x→3) f(x) = 7.
- Step 4: Defining f(3) = 7 makes the value match the limit.
Answer: Define f(3) = 7, since lim (x→3) f(x) = 7.
- Example 2
Find the constant
f(x) = kx + 1 for x < 2, and f(x) = x² − k for x ≥ 2. Find the value of k that makes f continuous for all x.
Show the solutionHide the solution
- Step 1: Each piece is a polynomial, so only x = 2 needs checking.
- Step 2: Left-hand limit: lim (x→2⁻) (kx + 1) = 2k + 1.
- Step 3: Right-hand limit and value: lim (x→2⁺) (x² − k) = f(2) = 4 − k.
- Step 4: For continuity, set them equal: 2k + 1 = 4 − k, so 3k = 3 and k = 1.
- Step 5: Check: with k = 1, the left piece gives 2(1) + 1 = 3 and the right piece gives 4 − 1 = 3.
Answer: k = 1
- Example 3
Trap: a jump can't be removed
g(x) = x + 1 for x < 1, and g(x) = 5 − x for x > 1. Can you choose a value for g(1) that makes g continuous at x = 1?
Show the solutionHide the solution
- Step 1: Left-hand limit: 1 + 1 = 2. Right-hand limit: 5 − 1 = 4.
- Step 2: The two one-sided limits are different, so lim (x→1) g(x) does not exist.
- Step 3: If you set g(1) = 2, the right side still ends at 4. If you set g(1) = 4, the left side still ends at 2. No single value meets both.
Answer: No. This is a jump discontinuity, so no value of g(1) makes g continuous at x = 1.
Common mistakes
- Plugging x = 2 into only one piece and solving that piece = 0. Set the two pieces equal to each other at the break point.
- Trying to remove a jump by redefining one point. It can't be done; the one-sided limits disagree.
- Giving the right constant but not checking it. Plug it back in to see that both pieces give the same height.
On the exam
- Find-the-constant questions are common in multiple choice. In free response they often pair with differentiability: one equation from continuity, one from matching slopes.
- For “define f(c) so that f is continuous,” the answer is always the value of the limit at c.
Connected topics
Videos
Check yourself
4 questions on 1.13 Removing Discontinuities. Pick an answer to see if you got it, and why.
Let f(x) = (x² + ax − 6)/(x − 2) for x ≠ 2, and f(2) = b, where a and b are constants. If f is continuous at x = 2, what are the values of a and b ?
Let f(x) = sin(2x)/(5x) for x ≠ 0, and f(0) = k. For what value of k is f continuous at x = 0 ?
Let f(x) = kx² − 1 for x ≤ 2, and f(x) = 3x + k for x > 2, where k is a constant. For what value of k is f continuous at x = 2 ?
Let f(x) = (e²ˣ − 1)/(eˣ − 1) for x ≠ 0. What value should f(0) be given so that f is continuous at x = 0 ?
0 of 4 answered