Removing Discontinuities: Study Guide (Calculus AB 1.13)

Course scope. This topic is shared by Calculus AB and Calculus BC. Everything here is examinable for both.

A note on notation

Limits are written in a compact form: lim (x → c) f(x) means “the limit as x approaches c of f(x)”. A minus or plus sign after c gives a one-sided limit: lim (x → c⁻) f(x) is the limit from the left, and lim (x → c⁺) f(x) is the limit from the right.

The idea: some breaks can be repaired

In Topic 1.11 you met the three conditions for f to be continuous at x = c:

1. f(c) is defined;
2. lim (x → c) f(x) exists;
3. the limit equals f(c).

A discontinuity happens when at least one condition fails. This topic asks a practical question: can you change the definition of f so that all three conditions hold?

You are only allowed to change one thing: the value of f at the single point c. You cannot change the values of f near c. That one rule decides everything.

• If the limit exists, you can set f(c) equal to it. Conditions 1, 2 and 3 then all hold. The discontinuity is removable.
• If the limit does not exist, changing one point does not help. Condition 2 still fails, whatever value you give f(c). The discontinuity is not removable.

So the test is simple: find the limit first. If it exists as a finite number, the discontinuity can be removed. If not, it cannot.

Removing a discontinuity: define or redefine f(c)

There are two versions of a removable discontinuity.

The value is missing. For example, f(x) = (x² − 4)/(x − 2) is undefined at x = 2. For x ≠ 2 it equals x + 2, so lim (x → 2) f(x) = 4. Define a new function with the missing value filled in:

• F(x) = (x² − 4)/(x − 2) for x ≠ 2
• F(2) = 4

F is continuous at 2. Notice that F is simply x + 2 for every x. Filling the hole gives back the whole line.

The value is in the wrong place. In the Topic 1.6 practice questions you met a function with f(x) = (x² − x − 6)/(x − 3) for x ≠ 3 and f(3) = 2. The limit at 3 is 5, but f(3) = 2. Condition 3 fails. Redefine f(3) = 5 and the function becomes continuous at 3. Nothing else changes.

The same idea works for any limit you can find, including limits from earlier topics. For example, g(x) = (sin 3x)/x is undefined at 0, and the result from Topic 1.8 gives lim (x → 0) g(x) = 3. Setting g(0) = 3 removes the discontinuity.

The method in four steps

1. Find where f is discontinuous. For a rational function, look for zeros of the denominator. For a piecewise function, look at the boundaries.
2. Find the limit there. Use the algebra of Topic 1.6, or one-sided limits.
3. Decide. If the limit exists as a finite number L, the discontinuity is removable. If the one-sided limits differ, or either is infinite, it is not.
4. Write the new definition. Set f(c) = L and state the function in full, including what happens at every other x.

Worked example 1: one hole, one asymptote

Question. Let f(x) = (x² + x − 12)/(x² − 5x + 6). Find every value of x where f is discontinuous. For each one, decide whether the discontinuity can be removed. If it can, say how.

1. Find the problem points. The denominator factors as x² − 5x + 6 = (x − 2)(x − 3). It is 0 at x = 2 and x = 3. f is undefined at both, so f is discontinuous at both.
2. Factor the top. x² + x − 12 = (x + 4)(x − 3).
3. Simplify. For x ≠ 2 and x ≠ 3, f(x) = (x + 4)(x − 3)/((x − 2)(x − 3)) = (x + 4)/(x − 2)
4. At x = 3. The simplified form has denominator 3 − 2 = 1, not 0, so substitute: lim (x → 3) f(x) = lim (x → 3) (x + 4)/(x − 2) = 7/1 = 7 The limit exists, so the discontinuity at x = 3 is removable. Define f(3) = 7.
5. At x = 2. In the simplified form the top is 2 + 4 = 6 and the bottom is 0: a nonzero number over 0. The function is unbounded near 2 (it goes to −∞ from the left and +∞ from the right). There is no finite limit, so this discontinuity is not removable. It is a vertical asymptote.

Answer. Removable at x = 3, by defining f(3) = 7. Not removable at x = 2.

Check. f(3.001) ≈ 6.994 and f(2.999) ≈ 7.006. Both are close to 7.

Writing it up. The new function is F(x) = (x² + x − 12)/(x² − 5x + 6) for x ≠ 2, 3, with F(3) = 7. Its domain still excludes x = 2. F is continuous at 3; nothing can make it continuous at 2.

Piecewise functions: making the pieces meet

A piecewise function uses different expressions on different intervals. Each piece is usually a polynomial, root, exponential or trig function, so each piece is continuous on its own interval (Topic 1.12). The only places that need checking are the boundaries, where one expression hands over to the next.

At a boundary x = c, three numbers must agree:

• the value the left expression gives at c (this is the limit from the left);
• the value the right expression gives at c (this is the limit from the right);
• f(c), given by whichever piece includes c (the one with ≤ or ≥).

If the left and right values match, the limit exists. Since f(c) comes from one of those two expressions, it matches as well. So in practice you write one equation: left expression at c = right expression at c. Then confirm f(c) agrees. If f(c) is given on a separate line, such as f(2) = m, it must equal that common value too.

If the function contains an unknown constant (a parameter), this equation lets you solve for it.

Worked example 2: solving for one parameter

Question. Find the value of k that makes f continuous at x = 2, where

• f(x) = kx + 6 for x < 2
• f(x) = kx² − 2 for x ≥ 2

1. Left value at 2. Substitute x = 2 into kx + 6: 2k + 6. This is lim (x → 2⁻) f(x).
2. Right value at 2. Substitute x = 2 into kx² − 2: 4k − 2. This is lim (x → 2⁺) f(x), and it is also f(2), because the second piece includes x = 2.
3. Set them equal. 2k + 6 = 4k − 2.
4. Solve. 8 = 2k, so k = 4.
5. Check all three values. Left: 2(4) + 6 = 14. Right: 4(4) − 2 = 14. f(2) = 14. All equal, so f is continuous at 2.

Answer. k = 4.

Interpretation. For x < 2 and x > 2, f is a polynomial, so it is continuous there. With k = 4 it is also continuous at 2, so f is continuous for all real x.

Worked example 3: two parameters, two boundaries

Question. Find the constants a and b that make g continuous for all real x, where

• g(x) = x + 1 for x < 1
• g(x) = ax² + b for 1 ≤ x ≤ 3
• g(x) = 4x − 2 for x > 3

1. Count unknowns and conditions. There are two unknowns, a and b, and two boundaries, x = 1 and x = 3. Each boundary gives one equation.
2. Boundary at x = 1. Left: 1 + 1 = 2. Right (middle piece, which includes 1): a(1)² + b = a + b. Equation: a + b = 2.
3. Boundary at x = 3. Left (middle piece, which includes 3): a(3)² + b = 9a + b. Right: 4(3) − 2 = 10. Equation: 9a + b = 10.
4. Solve simultaneously. Subtract the first equation from the second: 8a = 8, so a = 1. Then b = 2 − 1 = 1.
5. Check. The middle piece is x² + 1. At x = 1: left 2, middle 2. At x = 3: middle 10, right 10. Both boundaries join.
6. Conclude. Each piece is a polynomial, so it is continuous on its own interval. Both boundaries now join. So g is continuous for all real x.

Answer. a = 1 and b = 1.

When it is impossible, or there are two answers

The question for this topic often says “if possible”. Two outcomes need care.

No value works. Take f(x) = kx + 3 for x < 0 and f(x) = x² + 1 for x ≥ 0. At x = 0 the left piece gives k(0) + 3 = 3 for every k, and the right piece gives 1. The parameter disappears at the boundary, so the equation 3 = 1 has no solution. No k makes f continuous at 0. Say so clearly; that is the answer.

More than one value works. Take f(x) = x² + k² for x < 1 and f(x) = 5kx − 5 for x ≥ 1. The boundary condition is 1 + k² = 5k − 5, which rearranges to k² − 5k + 6 = 0, or (k − 2)(k − 3) = 0. So k = 2 or k = 3. Check both: with k = 2 the pieces give 5 and 5; with k = 3 they give 10 and 10. Give both values.

A jump cannot be removed by redefining one point. Suppose a function has lim (x → c⁻) f(x) = 2 and lim (x → c⁺) f(x) = 5. Setting f(c) = 2 makes the left side match but not the right. Setting f(c) = 5 does the reverse. The two-sided limit still does not exist, so f stays discontinuous at c. A jump can only be fixed by changing the pieces themselves, which is what a parameter does.

Using technology

You may meet this topic with or without a graphing calculator. Two cautions apply when you do use one:

• A plotted graph usually does not show a hole. The calculator samples points, and it almost never lands exactly on x = c. A smooth-looking curve is not proof of continuity.
• A table of values near c suggests a limit. It does not give the exact value you need for f(c). Use algebra to find the exact limit, and use the table only as a check.

Common misconceptions

• “Every discontinuity can be removed.” Only those where the limit exists. Jumps and vertical asymptotes cannot be removed by changing one value.
• “Removing a discontinuity means cancelling the factor.” Cancelling helps you find the limit. The removal itself is a statement about f(c): you must say what value f takes at c.
• “If f(c) is already defined, f is continuous at c.” f(c) must also equal the limit. A point in the wrong place still gives a discontinuity, and you remove it by redefining f(c).
• Setting a piece equal to 0 instead of setting the two pieces equal to each other at the boundary.
• Substituting the wrong x. The boundary value c goes into both expressions. Do not use a point from inside one interval.
• Forgetting the “both values” check. After solving, substitute back into both expressions. It catches sign slips, and it shows the examiner that the pieces meet.
• Dropping a solution. A quadratic condition can give two parameter values. Give both unless the question restricts k.
• Forcing an answer. If the equation has no solution, the correct answer is “no value works”. Do not invent one.
• Trusting a calculator graph. A plot that looks unbroken may still have a hole.

Where this leads

Topic 1.12 showed you how to find the intervals on which a function is continuous (previous study guide). This topic repairs the breaks that can be repaired. Next, Topic 1.14 studies the breaks that cannot: infinite limits and vertical asymptotes, like the one at x = 2 in Worked example 1. The same boundary condition returns in Unit 2, where a piecewise function must be continuous before it can be differentiable at a boundary. Students in either course can return to the Calculus AB roadmap or the Calculus BC roadmap.

Try the practice questions now, then use the revision notes and the checklist to consolidate.