Types of Discontinuities | AP Calculus Topic 1.10 Notes

1. Start with the three continuity conditions

A discontinuity is a failure of continuity at a point under consideration. In AP Calculus Topic 1.10, the main patterns are removable discontinuities, jump discontinuities and discontinuities associated with vertical asymptotes. Classify them using nearby behavior and the assigned function value.

At an interior point where both approaches are relevant, a function is continuous exactly when all three conditions hold:

The target value $f (a)$ is defined.
The two-sided limit $lim_{x \to a} f (x)$ exists as a finite real number.
The limit equals the assigned value:

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

To check the second condition, compute both one-sided limits. The first and third conditions then distinguish a finite limiting height from the value actually included at the target.

Use the definition rather than the pencil test

The “draw without lifting your pencil” idea is a useful visual intuition, not a complete test. A corner can be continuous, a small hole can be hidden by graph resolution, and endpoint continuity uses the approach available within the domain.

For a domain endpoint, use the appropriate one-sided continuity condition. A missing approach outside the domain does not by itself mean the endpoint is discontinuous. See Topic 1.11: continuity at a point and Topic 1.12: continuity over an interval.

2. The three main types at a glance

Removable: a hole or wrong point value

The finite two-sided limit exists, but the assigned value is missing or different. Defining that single value to equal the limit repairs continuity.

Jump: two finite heights

The left and right limits both exist as finite numbers and differ. Changing one point cannot make the nearby branches agree.

Infinite: unbounded one-sided behavior

At least one one-sided limit is positive or negative infinity. This supports a vertical asymptote; a single finite reassignment cannot remove that behavior.

These are the principal types emphasized in this topic. They are not an exhaustive list of every possible failure of a limit: bounded oscillation, for example, may have no one-sided limit and is not a finite jump.

3. Removable discontinuities: preserve the nearby limit

A removable discontinuity has a common finite approached value:

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

The assigned target value is either undefined or defined and different from that common value. In the defined case:

f (a) \neq L

A graph has an open point at the limiting height and may also have a filled point elsewhere at the same input. A common algebraic source is cancellation that leaves a function with a finite limit; another is a special target branch that differs from the nearby rule. To repair the function, assign the common finite limit at the target.

Example 1: a classic rational hole

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

At input two, both original numerator and denominator are zero, so the quotient is undefined. For nearby inputs different from two, factor and cancel:

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

lim_{x \to 2} \frac{x^{2} - 4}{x - 2} = lim_{x \to 2} (x + 2) = 4

The graph is the line with the input plus two as its output, except for an open point at $(2, 4)$ . The finite limit exists while the original target value is missing, so the discontinuity is removable.

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

This newly defined function is continuous at two. Cancellation by itself did not fill the original hole.

Example 2: a special value on a parabola

f (x) = {\begin{cases} x^{2} & if x \neq 3 \\ 5 & if x = 3 \end{cases}

lim_{x \to 3} f (x) = 9, f (3) = 5

The squared rule controls both nearby approaches; the isolated value five controls only the included point. Since the finite limit and the assigned value differ, the discontinuity is removable.

The open point

(3, 9)

marks the limiting height. The filled point

(3, 5)

records the assigned value. Redefining that value to nine repairs continuity.

4. Jump discontinuities: both sides finite and unequal

A jump requires two finite one-sided limits with different values:

lim_{x \to a^{-}} f (x) = L_{1}, lim_{x \to a^{+}} f (x) = L_{2}

L_{1} \neq L_{2}

The target value may exist or be missing. A piecewise function, a step function or a change of rules can produce a jump, but different formulas alone do not prove one. Evaluate the side limits. Redefining one value does not alter either nearby approach, so it cannot remove a jump.

Example 3: a boundary with different approached heights

h (x) = {\begin{cases} x - 1 & if x < 1 \\ 2 - x & if x \geq 1 \end{cases}

lim_{x \to 1^{-}} h (x) = lim_{x \to 1^{-}} (x - 1) = 0

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

h (1) = 1

The finite side limits differ, so there is a jump at input one. Its size is the absolute difference of the limiting heights:

| 1 - 0 | = 1

The left branch has an open point at

(1, 0)

; the right branch includes

(1, 1)

. No vertical connector is part of the graph.

Example 4: a downward step

f (x) = {\begin{cases} 2 & if x < 0 \\ - 3 & if x \geq 0 \end{cases}

lim_{x \to 0^{-}} f (x) = 2, lim_{x \to 0^{+}} f (x) = - 3

| 2 - (- 3) | = 5

The assigned value at zero is negative three. It does not make the different nearby heights agree; the two-sided limit does not exist as a finite value.

5. Infinite discontinuities and vertical asymptotes

An infinite discontinuity is associated with at least one one-sided unbounded approach:

lim_{x \to a^{-}} f (x) = \pm \infty, or lim_{x \to a^{+}} f (x) = \pm \infty

Each sign in a specific problem must be determined; the symbol with both signs is a list of possibilities, not one common output. The graph may rise or fall without bound near the vertical line at the target. A dashed line is a guide that should be supported by the function's behavior.

Rational denominator zeros that remain after complete simplification often create poles. Logarithms can also have a vertical asymptote at a real-domain boundary. Check the actual domain and side limits rather than declaring an asymptote merely because a denominator becomes zero.

The function value does not control an asymptote

The original rational formula is usually undefined at its pole, but a piecewise definition can assign a finite value at that same input. The one-sided unbounded behavior remains. Thus a function with a vertical asymptote can still have an isolated defined point on the asymptote line.

Example 5: both sides tend to positive infinity

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

The denominator is positive at every nearby input other than two and approaches zero from above; the numerator is positive. Therefore:

lim_{x \to 2^{-}} \frac{3}{{(x - 2)}^{2}} = \infty, lim_{x \to 2^{+}} \frac{3}{{(x - 2)}^{2}} = \infty

lim_{x \to 2} \frac{3}{{(x - 2)}^{2}} = \infty

This notation is valid: it states unbounded growth, not a finite real limit. The graph has two disconnected branches, not one continuous curve across the asymptote.

Both branches rise without bound near input two. The dashed line marks the vertical asymptote. Curves continuing beyond the upper plotting window are clipped by the viewport, not endpoints of the function.

Example 6: the sides have opposite infinite directions

q (x) = \frac{1}{x - 3}, x \neq 3

lim_{x \to 3^{-}} \frac{1}{x - 3} = - \infty

lim_{x \to 3^{+}} \frac{1}{x - 3} = \infty

The denominator is small negative on the left and small positive on the right. There is a vertical asymptote at input three, but no single common two-sided infinite limit. State the two directions explicitly.

6. Compare the three types

Compare limiting behavior, target membership and removability
Feature	Removable	Jump	Infinite behavior
Nearby limits	Both sides approach the same finite value.	Both one-sided limits are finite and different.	At least one relevant side tends to positive or negative infinity.
Two-sided limit	Exists as a finite real number.	No common two-sided finite limit.	No finite real limit; a common signed infinite limit can exist when both sides agree in that direction.
Assigned target value	Missing or different from the finite limit.	May be defined or undefined.	May be undefined in the original formula or separately assigned a finite value.
Graph	A hole at the limiting height, possibly with a separate filled point.	Distinct approached heights; no invented vertical connector.	A vertical asymptote supported by unbounded side behavior.
Single-point repair	Yes: assign the finite limiting value.	No: the nearby side limits remain unequal.	No: a finite reassignment cannot remove unbounded nearby behavior.
Typical source	A fully canceled obstruction with a finite remaining limit, or an isolated changed value.	A piecewise mismatch or a step.	A pole or a logarithmic domain boundary, verified through limits.
Terminology	Also called a point discontinuity in many texts.	Also described as a step discontinuity.	Use “infinite discontinuity” or “vertical asymptote”; broader terms such as “essential” vary between texts.

7. A reliable identification procedure

1. Keep the original domain
List denominator exclusions, branch boundaries and other domain restrictions before simplifying.
2. Examine each relevant side
Find the one-sided limits at an interior target. At an endpoint, respect the available domain approach.
3. Check unbounded behavior
If a side tends to a signed infinity, identify the vertical asymptote and state each known side separately.
4. Compare finite side limits
If both are finite but unequal, there is a jump. If a side limit fails through oscillation, do not automatically call it a jump.
5. If both finite sides agree, inspect the value
A missing or different target value gives a removable discontinuity. An equal defined value gives continuity.
6. State the classification with evidence
Name the input, the type and whether a single-point repair is possible. If the data are insufficient, say what remains unknown.

For example, the bounded function sine of the reciprocal input has no limit at zero because it keeps oscillating. It is neither a removable discontinuity nor a finite jump nor a vertical-asymptote case. This is why “not infinite” followed by “not equal” is not a complete classification rule.

Use Topic 1.9: connecting representations to distinguish a numerical or graphical suggestion from a justified limit calculation.

8. Algebraic techniques and the limits of shortcuts

For a possible removable discontinuity

Factor the rational numerator and denominator.
Record the original excluded inputs.
Cancel common factors only where they are nonzero.
Check whether the reduced expression has a finite limit at the excluded input.
If so, compare that limit with the assigned value and locate the hole.

A cancellation alone is insufficient. Consider:

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

A factor cancels, but a denominator factor remains at the same input. The nearby behavior is unbounded, so the point is not a removable hole. Complete the simplification and limit analysis before classifying it.

For a possible jump

Choose the branch valid below the target and the branch valid above it. Evaluate their limits separately, checking any domain or discontinuity inside those branches. A jump occurs when both results are finite and unequal. The target value comes from the branch that includes the equality sign.

For a possible vertical asymptote

Locate candidates, simplify while retaining exclusions, then analyze signs and magnitude on both sides. For a rational function reduced to relatively prime polynomials, a real denominator zero creates a pole; the side signs depend on multiplicity and nearby numerator signs. Logarithmic examples require domain-aware one-sided analysis.

Review algebraic manipulation of limits and infinite limits and vertical asymptotes for further worked steps.

9. Example 7: one function with two different discontinuities

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

Factor the original numerator and denominator:

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

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

The original domain excludes both two and three. At every allowed input:

\frac{x^{2} - 9}{x^{2} - 5 x + 6} = \frac{(x - 3) (x + 3)}{(x - 2) (x - 3)} = \frac{x + 3}{x - 2}

At input three: a finite removable hole

lim_{x \to 3} \frac{x^{2} - 9}{x^{2} - 5 x + 6} = \frac{6}{1} = 6

The reduced expression is continuous at three, but the original quotient is undefined there. Thus the hole is at $(3, 6)$ . Defining the target value to be six removes this discontinuity.

At input two: an infinite discontinuity

Near two the reduced numerator is positive and tends to five. Its denominator tends to zero with the input's side:

lim_{x \to 2^{-}} \frac{x^{2} - 9}{x^{2} - 5 x + 6} = - \infty

lim_{x \to 2^{+}} \frac{x^{2} - 9}{x^{2} - 5 x + 6} = \infty

The vertical asymptote is at input two. A single-point definition cannot remove it. The function has different types at different inputs: removable at three and infinite at two.

The graph is split at both original exclusions. The open point

(3, 6)

remains even though the reduced formula is defined there. The dashed vertical line marks the pole at input two.

10. Memory aids and recognition patterns

RJI: repairable, jump, infinity

R — Removable: the finite nearby limit can repair the target value.
J — Jump: the two finite heights disagree.
I — Infinite: at least one side is unbounded in a definite signed direction.

“Hole, step, wall” can help remember the pictures, but use limits to justify the classification. A removable discontinuity may include a wrong filled point as well as a hole, and an asymptote is not an actual wall in the graph.

Recognize a candidate, then verify

$\frac{0}{0}$ after substitution identifies an indeterminate form. Try a valid method; it does not by itself establish a removable discontinuity.
A nonzero limiting numerator with denominator tending to zero suggests unbounded behavior. Check the side, sign and domain.
A piecewise boundary needs a comparison of the side limits, not an automatic “jump” label.
A canceled factor needs a finite remaining limit before the excluded point is called a hole.
Standard floor and ceiling functions jump at every integer; their side limits are finite and differ by one.
An absolute value alone does not create a discontinuity. The absolute-value function is continuous at zero, though it has a corner there.

lim_{x \to 0} | x | = 0

Distinguish a continuity question from a differentiability question. A point can be continuous but have different one-sided slopes.

11. Common mistakes and precise corrections

Calling every failure removable. A single-point repair requires a finite common nearby limit.
Checking only one side at an interior point. A shared two-sided conclusion must account for both approaches.
Confusing the point value with the limit. Read open and filled markers separately.
Losing the original domain during cancellation. A reduced formula may be defined at a point the original excludes.
Calling every piecewise boundary a jump. Matching sides and a matching target value can give continuity.
Rejecting valid infinite-limit notation. Writing a limit equal to positive infinity is standard notation for unbounded growth, while infinity is not a finite real function value.
Assuming a vertical asymptote requires an undefined target value. An artificial isolated value can coexist with the same unbounded nearby behavior.
Treating the three main types as exhaustive. Oscillation may require a separate description.
Assigning conflicting principal labels at one input. A finite common limit, unequal finite side limits and a signed infinite side limit are distinct conditions. One function can have different classifications at different inputs.

Terminology such as “essential discontinuity” has broader or different meanings in different texts. Explicit descriptions of the side limits avoid that ambiguity.

12. Write an AP Calculus classification argument

Use the continuity definition and the actual limit behavior to support the classification. A useful written response has four parts:

Location: identify the candidate input and original domain restriction or branch boundary.
Limits: compute the relevant one-sided limits or explain why a side fails to settle.
Target value: state whether it is defined and, when a finite common limit exists, compare the two.
Conclusion: name the discontinuity type and explain whether redefining one value would repair continuity.

For a graph, mark open and filled points correctly and separate branches at jumps or asymptotes. For a rational formula, show enough factoring to support every original exclusion. For numerical data, distinguish a suggested pattern from a conclusion justified by supplied structure.

Typical questions

Identify and classify all discontinuities of a formula or graph.
Determine whether a function is continuous at a specified input and justify the answer.
Locate a removable, jump or infinite discontinuity.
Find a value that removes a discontinuity, if possible.
Sketch the function with correct holes, included points and asymptotes.

Precise locations, correct side-limit calculations, appropriate classifications and a clear removability statement make the reasoning reviewable. Any scoring details depend on the particular question; there is no universal point allocation for every Topic 1.10 response.

13. Practice problems with complete explanations

Determine whether the function is continuous or has a discontinuity at the requested input. Include the side limits, target value and removability decision before opening each solution.

1. A difference-of-squares quotient at input four

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

Check the classification and reasoning

The original value at four is undefined. Factor the numerator and cancel the nonzero nearby factor:

\frac{x^{2} - 16}{x - 4} = x + 4

The nearby limit is eight, so this is removable, with a hole at $(4, 8)$ . Defining the value to be eight repairs it.

2. A piecewise function at input two

g (x) = {\begin{cases} 3 x & if x < 2 \\ x + 4 & if x \geq 2 \end{cases}

Check the classification and reasoning

lim_{x \to 2^{-}} 3 x = 6, lim_{x \to 2^{+}} (x + 4) = 6

The assigned value is also six, so the function is continuous at two. There is no discontinuity to classify, despite the change in formulas.

3. A rational function at input five

h (x) = \frac{2 x + 1}{x - 5}

Check the classification and reasoning

The numerator tends to eleven, which is positive. The denominator changes from negative to positive across five:

lim_{x \to 5^{-}} \frac{2 x + 1}{x - 5} = - \infty, lim_{x \to 5^{+}} \frac{2 x + 1}{x - 5} = \infty

There is an infinite discontinuity and vertical asymptote at input five. No finite single-point assignment removes it.

4. A rational quotient at input one

p (x) = \frac{x^{2} - 1}{x^{2} + x - 2}

Check the classification and reasoning

\frac{x^{2} - 1}{x^{2} + x - 2} = \frac{x + 1}{x + 2}

The original domain excludes one and negative two. At the requested input one, the reduced denominator is nonzero and the finite limit is $\frac{2}{3}$ . Hence the discontinuity there is removable, with a hole at $(1, \frac{2}{3})$ . The other excluded input is a separate vertical-asymptote candidate.

14. Quick reference card

Identify the limit behavior before applying the label
Type	Limit evidence	Point value	Single-point repair
Removable	$lim_{x \to a} f (x)$ is a finite common value.	Undefined, or defined and different from that limit.	Assign the finite limit at the target.
Jump	Two finite one-sided limits exist and differ.	May be included on either branch, separately assigned or absent.	Impossible: changing the target value cannot change the side limits.
Infinite	At least one relevant side tends to a signed infinity.	May be absent or separately assigned.	Impossible with a finite point reassignment.
Continuous	The relevant finite limit equals the assigned value.	Defined and matches the limit.	No repair is needed.
Other behavior	For example, a side oscillates without approaching a finite value or a signed infinity.	Inspect separately.	Do not force it into the finite-jump classification.

15. Why discontinuity classification matters

This topic prepares you to check continuity at a point or over an interval, remove finite holes and verify hypotheses for the Intermediate Value Theorem. A correct classification explains what breaks and whether a single reassignment can fix it.

Differentiation: differentiability at an interior point implies continuity there, so a discontinuity rules out a finite derivative at that point. Continuity alone does not guarantee a derivative.
Applications: side behavior and domain restrictions help interpret models with switching rules, steps or unbounded values.
Integration: continuity is a useful sufficient condition, not a requirement for every definite integral. Bounded functions with finitely many jumps can be integrable; unbounded cases need an appropriate improper-integral analysis, which may converge or diverge.

After reviewing the remaining unit topics, use the Unit 1 practice test collection to apply the classifications across formulas, graphs and numerical information.

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