Removing Discontinuities | AP Calculus Topic 1.13 Notes

1. What makes a discontinuity removable?

A removable discontinuity occurs when a function approaches a common finite value at an input, but its assigned value there is missing or different. You can restore continuity by defining or redefining that one value to match the limit.

lim_{x \to a^{-}} f (x) = lim_{x \to a^{+}} f (x) = L

The original value $f (a)$ is either undefined or, if defined, unequal to $L$ . If it already equals the finite limit, the function is continuous there and no repair is needed.

Finite is essential

A jump, a vertical asymptote or oscillation without a limit cannot be repaired by assigning one finite value. Equal infinite directions also do not supply a finite value for a continuous extension.

At an interior point, compare both sides. At a domain endpoint, use the approach available within the domain when discussing a one-sided continuous extension. Review Topic 1.11: the definition of continuity.

2. Read the hole and the repaired point

An open circle at $(a, L)$ shows the finite limiting height that is missing from the graph. A separate filled point at the same input may show an assigned value that disagrees with the limit. Follow the nearby curve before reading the included point.

Before repair, the original rational function excludes input two and has a hole at

(2, 4)

. The curve is clipped by the plotting window, not restricted to these endpoints.

After defining the value at two as four, the same nearby line includes

(2, 4)

. The extension changes only that input.

Filling a hole restores continuity; it does not automatically make every function smooth or differentiable. The repaired function's other inputs still need their own continuity checks. See reading limits from graphs.

3. Why holes and mismatched values occur

Rational expressions with common factors: a denominator excludes the target even when algebraic simplification produces a finite nearby limit. The original exclusion remains until a value is explicitly added.
Piecewise definitions: neighboring rules may approach the same finite value while the target-only branch supplies another value or omits it.
A deliberately excluded input: a rule may be specified only away from the target even though it has a finite limit there.

Direct substitution giving $\frac{0}{0}$ identifies an indeterminate form, not a function value and not proof that the discontinuity is removable. Complete the limit calculation. Likewise, cancellation alone does not guarantee a hole if a remaining denominator still causes unbounded behavior.

An expression can have several excluded inputs with different classifications. Preserve the full original domain and analyze each input separately.

4. The three-step repair and why it works

Verify that a repair is possible.
Find a common finite nearby limit and check that the original target value is missing or mismatched. If the required finite limit does not exist, stop: a target-only reassignment cannot work.
Calculate the exact limit.
Factor and cancel valid nonzero factors, use another appropriate limit method or evaluate the neighboring piecewise rules. Keep every original exclusion.
Define the extension and verify it.
Keep the original rule at its other allowed inputs and assign the finite limit at the target. Then confirm the assigned value, nearby limit and equality.

If the original domain is $D$ and the finite nearby limit at the target is $L$ , define the extended function by:

g (x) = {\begin{cases} f (x) & if x \in D, x \neq a \\ L & if x = a \end{cases}

The extended domain is the original domain with the target included. If the target was already in the domain, this replaces its old value. It does not automatically add any other excluded inputs.

lim_{x \to a} g (x) = lim_{x \to a} f (x) = L = g (a)

The displayed equality explains the repair: limits depend on nearby values, so changing only the target does not alter the limit. The value chosen for continuity is unique; any different assignment would still disagree with that same limit.

5. Rational functions: factor, cancel, evaluate, define

For a rational expression producing an indeterminate form, factor both numerator and denominator. Cancel common factors only where they are nonzero. Do not cancel separate terms in a sum. Evaluate the remaining expression at the target only if its continuity and domain justify substitution.

The memory aid FCE means factor, cancel, evaluate. Add one final action: define the missing or corrected value. Algebraic equivalence away from the target does not by itself change the original function.

Counterexample: partial cancellation leaves a pole

\frac{x - 1}{{(x - 1)}^{2}} = \frac{1}{x - 1}, x \neq 1

Although the original substitution produces zero over zero and one factor cancels, the remaining quotient is unbounded near one. This discontinuity is not removable. For polynomials in a rational expression, complete cancellation of the target factor from the denominator leaves a nonzero target denominator; otherwise analyze the remaining zero and signs.

Other valid limit procedures, such as rationalizing a radical expression, can also reveal finite limiting values. Use Topic 1.6: algebraic manipulation and select a method suited to the expression.

6. Three worked rational-function examples

Example 1: a classic removable hole

f (x) = \frac{x^{2} - 4}{x - 2}, x \neq 2

The denominator excludes two. Factoring the difference of squares gives:

x^{2} - 4 = (x - 2) (x + 2)

\frac{x^{2} - 4}{x - 2} = \frac{(x - 2) (x + 2)}{x - 2} = x + 2, x \neq 2

lim_{x \to 2} f (x) = lim_{x \to 2} (x + 2) = 4

The original value at two is missing. Define the extension explicitly:

g (x) = {\begin{cases} \frac{x^{2} - 4}{x - 2} & if x \neq 2 \\ 4 & if x = 2 \end{cases}

The new value and finite limit both equal four, so the extension is continuous at two. In fact it is the line with output equal to input plus two at every real input.

Example 2: a repeated numerator factor

h (x) = \frac{x^{2} - 4 x + 4}{x - 2}, x \neq 2

x^{2} - 4 x + 4 = {(x - 2)}^{2}

\frac{x^{2} - 4 x + 4}{x - 2} = \frac{{(x - 2)}^{2}}{x - 2} = x - 2, x \neq 2

lim_{x \to 2} h (x) = 0

H (x) = {\begin{cases} \frac{x^{2} - 4 x + 4}{x - 2} & if x \neq 2 \\ 0 & if x = 2 \end{cases}

The extension agrees everywhere with the polynomial rule input minus two. It is therefore continuous at every real input. A repair value can be zero; what matters is the finite limit, not whether the value is positive or nonzero.

Example 3: a hole and a vertical asymptote

f (x) = \frac{x^{2} - 9}{x^{2} - 5 x + 6}

Factor both polynomials and record both original exclusions:

x^{2} - 9 = (x - 3) (x + 3)

x^{2} - 5 x + 6 = (x - 2) (x - 3)

\frac{x^{2} - 9}{x^{2} - 5 x + 6} = \frac{x + 3}{x - 2}, x \neq 2, x \neq 3

At three: the reduced denominator is nonzero, so the finite limit is six. This is the removable hole.

lim_{x \to 3} f (x) = \frac{6}{1} = 6

At two: the remaining numerator approaches five while the denominator approaches zero with opposite signs.

lim_{x \to 2^{-}} f (x) = - \infty

lim_{x \to 2^{+}} f (x) = \infty

No finite reassignment at two can remove the vertical asymptote. Extend only at three:

G (x) = {\begin{cases} \frac{x^{2} - 9}{x^{2} - 5 x + 6} & if x \neq 2, x \neq 3 \\ 6 & if x = 3 \end{cases}

The extension's domain still excludes two. On that domain it equals the reduced quotient, including at three. Conclusion: only the hole at three is repaired; the pole at two remains.

The original graph has an open point at

(3, 6)

and a vertical asymptote at input two. The repair fills that hole and leaves the asymptote unchanged. Curves are split at both original exclusions.

7. Piecewise functions: value parameters and branch parameters

At an interior boundary, compare the finite side limits with the actual value selected by the branch containing the equality sign:

lim_{x \to a^{-}} f (x) = lim_{x \to a^{+}} f (x) = f (a)

A parameter used only at the target can select the included value. It cannot change the nearby side limits. A solution exists only when those limits already agree finitely.
A parameter inside a neighboring rule can change the side limit itself. Solve for matching limits and then verify the included value and domain restrictions.

Evaluate each side with its own neighboring branch. If the finite side limits differ, the boundary is a jump. If a side is unbounded or fails to settle, describe that behavior separately rather than labeling every failure a jump.

For continuity on an entire interval, also check other branch boundaries and each included endpoint. A successful local repair does not prove global continuity. Review Topic 1.12: interval continuity.

8. Three worked parameter examples

Example 4: no target value can repair unequal limits

f (x) = {\begin{cases} 3 x + 2 & if x < 1 \\ k & if x = 1 \\ x^{2} + 1 & if x > 1 \end{cases}

lim_{x \to 1^{-}} f (x) = lim_{x \to 1^{-}} (3 x + 2) = 5

lim_{x \to 1^{+}} f (x) = lim_{x \to 1^{+}} (x^{2} + 1) = 2

The finite side limits are different. The parameter controls only the value at one, so it cannot change either approach. No value of the parameter makes this function continuous at one. This is a jump, not a removable discontinuity.

Example 5: change a branch parameter to make the join agree

g (x) = {\begin{cases} k x + 1 & if x < 4 \\ x^{2} - 7 & if x \geq 4 \end{cases}

lim_{x \to 4^{-}} g (x) = 4 k + 1

lim_{x \to 4^{+}} g (x) = g (4) = 4^{2} - 7 = 9

Set the left-hand expression equal to nine and solve:

4 k + 1 = 9

4 k = 8 \Rightarrow k = \frac{8}{4} = 2

At this parameter value both side limits and the included value equal nine. Both branches are polynomials with no other exceptional inputs, so the resulting function is continuous at every real input. Here choosing the parameter changes a whole branch; it is a different task from merely redefining one point of a fixed function.

Example 6: choose the missing value in a rational piecewise rule

f (x) = {\begin{cases} \frac{x^{2} - 9}{x - 3} & if x \neq 3 \\ a & if x = 3 \end{cases}

\frac{x^{2} - 9}{x - 3} = \frac{(x - 3) (x + 3)}{x - 3} = x + 3, x \neq 3

lim_{x \to 3} f (x) = 6

f (3) = a = 6

The unique choice is six. The resulting function equals the continuous line with output input plus three everywhere, including at three.

9. Removable versus non-removable behavior

A finite common limit is the deciding condition
Feature	Removable case	No finite single-point repair
Nearby limit	A common finite limit exists.	No common finite limit: unequal finite sides, unbounded behavior or other failure.
Assigned value	Missing, or defined and different from that limit.	May be defined or missing; either way it cannot fix the nearby behavior.
Graph	A hole at the finite approached height, possibly with a misplaced filled point.	A jump, vertical asymptote or oscillation without a limit.
Repair	Assign the common finite limit at the target.	Changing only the target cannot change the side behavior.
Example	$\frac{x^{2} - 1}{x - 1}$ at input one.	$\frac{1}{x}$ at input zero.

A limit can be written equal to infinity as valid notation for unbounded growth, but that is not a finite number to assign for continuity. A finite point value also can coexist with an asymptote; the nearby unbounded behavior remains.

10. When a single-point repair cannot work

Unequal finite side limits

A target value cannot equal two different approached heights at once. Step functions, such as the floor function at integer boundaries, provide familiar jumps.

A vertical asymptote

lim_{x \to 2^{-}} \frac{1}{x - 2} = - \infty

lim_{x \to 2^{+}} \frac{1}{x - 2} = \infty

Neither side gives a finite height for an extension, and the directions are opposite. Do not combine them into a single two-sided signed infinity. See Topic 1.14: infinite limits and vertical asymptotes.

Oscillation without a limit

\sin (\frac{1}{x}), x \to 0

This function is bounded, but its values keep reaching different heights arbitrarily close to zero. Boundedness alone does not give a limit. No target value can make a continuous extension there.

11. Recognition patterns and exam reasoning

Recognize a candidate, then verify it. A zero-over-zero form suggests a limit calculation; it does not establish a hole. A common factor may help simplify, but inspect what remains. Matching finite side limits with a missing or mismatched value give the actual removable condition.

Factor completely: show enough algebra to justify cancellation and retain original exclusions.
Use both neighboring rules: establish the relevant one-sided limits separately at piecewise joins.
State the finite limit: explain why it exists rather than simply assigning a plausible number.
Write the extension clearly: specify the new target value and retain the original rule at its other allowed inputs.
Verify the result: the new value is defined, the relevant limit is finite and they agree.
Label a sketch carefully: use an open marker for the excluded original point and an included marker for the extension.

Typical questions

Find a parameter that makes a function continuous at a specified input.
Identify all discontinuities and remove those that are removable.
Decide whether a discontinuity is removable and justify the conclusion.
Redefine a function to make it continuous at a target.
Determine where the original or extended function is continuous.

Write a reason suited to the question, including restrictions and calculations that support your conclusion. There is no universal response format that guarantees a particular exam score.

12. Common mistakes and better checks

Assuming every discontinuity is removable. A finite common nearby limit is essential.
Equating cancellation with a filled hole. Cancellation only gives an equivalent rule at allowed inputs; define the new value explicitly.
Dropping other exclusions. The mixed rational example still excludes two after repairing three.
Canceling terms instead of factors. Factor first; do not remove individual terms from sums.
Substituting into an undefined original quotient. Use a justified limiting expression to find the repair value.
Checking only one side at an interior join. Both finite side limits and the included value must agree.
Trying to repair a jump or pole with a finite value. Changing one point leaves the neighboring behavior unchanged.
Confusing a branch parameter with a point-only parameter. Determine which parts of the function the parameter changes.
Promising smoothness or global continuity. A repair proves the relevant continuity claim, not every differentiability or interval claim.

Use Topic 1.10: types of discontinuities to classify the failure before deciding how to repair it.

13. Four practice problems with complete solutions

All four original problems are retained. Establish the finite nearby limit, if it exists, before assigning a value or choosing a parameter.

1. Remove the hole at input four

f (x) = \frac{x^{2} - 16}{x - 4}

Check removability and the complete solution

\frac{x^{2} - 16}{x - 4} = \frac{(x - 4) (x + 4)}{x - 4} = x + 4, x \neq 4

lim_{x \to 4} f (x) = 8

The original value at four is missing, while the common finite limit is eight. Keep the original quotient at every other input and define the extended value at four to be eight. This repairs the removable hole.

2. Decide whether input one is repairable

g (x) = \frac{x + 3}{x - 1}

Check removability and the complete solution

lim_{x \to 1^{-}} g (x) = - \infty

lim_{x \to 1^{+}} g (x) = \infty

The numerator approaches four and the denominator approaches zero with opposite signs. There is no finite two-sided limit. The vertical asymptote is not removable by a finite target reassignment.

3. Choose the target value at input three

h (x) = {\begin{cases} x^{2} & if x < 3 \\ k & if x = 3 \\ 6 x - 9 & if x > 3 \end{cases}

Check removability and the complete solution

lim_{x \to 3^{-}} h (x) = 9 = lim_{x \to 3^{+}} h (x)

h (3) = k = 9

The left rule approaches nine, and the right rule approaches eighteen minus nine, also nine. The parameter controls only the included value, so nine is the unique choice for continuity at three.

4. Remove the hole in a cubic quotient at two

p (x) = \frac{x^{3} - 8}{x - 2}

Check removability and the complete solution

x^{3} - 8 = (x - 2) (x^{2} + 2 x + 4)

\frac{x^{3} - 8}{x - 2} = x^{2} + 2 x + 4, x \neq 2

lim_{x \to 2} p (x) = 12

The original quotient excludes two. Keep its other values and assign twelve at that input; the extension then equals a polynomial everywhere and is continuous.

14. Quick reference: limit first, value second

A removable discontinuity has one unique continuous extension value
Step	Action	Evidence
Identify	Check the original value and the relevant side limits.	A missing or mismatched value with a common finite nearby limit.
Calculate	Use valid algebra or the neighboring branch rules.	$lim_{x \to a} f (x) = L$ .
Extend	Keep other original values and assign the finite limit at the target.	$g (a) = L$ .
Verify	Compare the extended value with its finite limit.	$lim_{x \to a} g (x) = g (a)$ .
If no finite common limit	A single finite target reassignment cannot work.	Explain the jump, unbounded behavior or other failure instead.

A successfully repaired point may still share its function with other non-removable discontinuities. State precisely where the original and extended functions are defined and continuous.

15. Connections and the next topic

Removing a discontinuity combines limits, algebra and the definition of continuity. It also reinforces the difference between a formula and a function with a specified domain.

Topics 1.10–1.12: classify discontinuities and use point or interval continuity to verify a repair.
Topic 1.14: signed infinite limits explain vertical asymptotes that a finite reassignment cannot remove.
Topic 1.16, Intermediate Value Theorem: continuity on the full closed interval is needed before invoking its guarantee.
Derivatives: differentiability implies continuity, but filling a hole does not automatically prove differentiability.
Unit 5 applications: check domains and the hypotheses of extrema and optimization results rather than assuming continuity from a model's appearance.
Modeling: a mathematically justified continuous extension differs from guessing a missing data value; finite observations alone do not prove an exact limit.

Previous: Topic 1.12 — Interval continuity →Next: Topic 1.14 — Infinite limits and vertical asymptotes →All Unit 1 study notes →Unit 1 practice test collection →

For a repair question, establish the finite limit first, state the extension explicitly and use the continuity definition to verify the result.

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