Continuity at a Point | AP Calculus Topic 1.11 Notes

1. What continuity at a point means

A function is continuous at a point when its nearby outputs approach the value actually assigned at that input. At an interior point, both approaches must lead to the same finite height, and the included point must lie at that height.

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

The equation includes two existence requirements: the function value must be defined as a real number, and the limit must exist as a finite real number. A missing value, two different approached heights or unbounded behavior prevents ordinary continuity there.

Use the pencil test as intuition

Drawing through a point without lifting a pencil is a helpful picture, but the three conditions give the actual test. A continuous function can have a corner: continuity does not require a smooth curve or a derivative.

This topic focuses on justifying a conclusion at a specified input. Checking one point does not establish continuity on an entire interval. Review limit notation if you need to separate nearby behavior from an assigned value.

2. The formal three-condition definition

For an interior point $x = c$ where both approaches are relevant, continuity requires all three conditions.

1. The value is defined

$f (c)$ must be a real number. Use the actual rule or the piecewise branch containing the target, rather than an expression obtained only by taking a limit.

2. A finite limit exists

The left and right limits must both be finite and equal. This describes nearby behavior independently of the target value.

3. The limit matches the value

The common approached height must equal the assigned height. A correct limit and a defined point are not enough if they disagree.

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

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

What can break each condition?

Undefined value: division by zero, a negative radicand in a real square root, a nonpositive logarithm argument or another domain exclusion.
No finite limit: unequal finite side limits, unbounded behavior or oscillation that does not settle.
Mismatch: a special point value differs from the finite nearby limit. If the value is missing, the equality cannot be satisfied either.

“The limit exists” in this continuity test means exists as a finite real number. Equal signs of infinity do not give a finite value. Standard infinite-limit notation is valid for describing unbounded growth, but infinity is not a real function value establishing continuity.

The compact equation includes existence

The equality between the limit and the function value summarizes the definition only when both quantities exist in the required finite sense. Do not treat an undefined expression as a number.

3. Apply the three-condition test step by step

Read the target and domain.
Identify restrictions and which piecewise branch includes the input. Equality belongs to the branch whose condition actually contains it.
Find the assigned value.
Substitute into the applicable rule. A missing value already prevents continuity as originally defined; continue checking the limit if classification or a repair is requested.
Find the nearby limit.
Use the rule immediately to the left for the left limit and the rule immediately to the right for the right limit. A branch defined only at the target does not control nearby approaches.
Compare and justify.
When the finite side limits match, compare their common value with the assigned value. State which condition fails or explain why all three hold.

For a polynomial or another known continuous function at an allowed input, a continuity theorem justifies direct substitution. A suspicious rational expression or piecewise boundary needs the additional algebra or side-limit work. Review algebraic manipulation for limits.

A numerical table may suggest a limit but does not by itself prove exact continuity. Calculator graphs can hide holes or small jumps. Use the definition and the structure supplied in the problem.

4. Five complete worked examples

Example 1: a polynomial at input two

f (x) = 2 x^{2} + 3 x - 1

Condition one: evaluate the assigned value.

f (2) = 2 {(2)}^{2} + 3 (2) - 1 = 13

Condition two: polynomials are continuous at every real input, so direct substitution gives the limit.

lim_{x \to 2} (2 x^{2} + 3 x - 1) = 13

Condition three and conclusion: both quantities are thirteen, so the function is continuous at two.

lim_{x \to 2} f (x) = 13 = f (2)

Example 2: an excluded input and a removable hole

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

Condition one fails: the original denominator is zero at two, so the function has no assigned value there. Substitution produces the indeterminate form below; it is not a value of the function.

\frac{0}{0}

Condition two: factor and cancel only for inputs different from two.

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

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

Condition three and conclusion: the finite limit exists, but the original target value is missing. The equality required for continuity cannot hold. This is a removable hole at $(2, 4)$ . Assigning four there would give a continuous extension; cancellation alone does not redefine the original function.

The original rational expression approaches the open point

(2, 4)

. There is no included point at that input.

Example 3: different formulas that join continuously

h (x) = {\begin{cases} x^{2} & if x < 2 \\ 4 & if x = 2 \\ x + 2 & if x > 2 \end{cases}

Condition one: the equality branch supplies the value.

h (2) = 4

Condition two: the neighboring branches determine the side limits.

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

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

Condition three and conclusion: both side limits and the assigned value agree. The function is continuous at two even though its formula changes there.

lim_{x \to 2} h (x) = 4 = h (2)

The branches meet at the included point

(2, 4)

. A corner does not prevent continuity; it concerns differentiability instead.

Example 4: a defined value with unequal side limits

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

Condition one: the right branch includes the boundary.

p (1) = 2 - 1 = 1

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

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

Condition two fails: the finite side limits differ, so the two-sided limit does not exist. The function is not continuous at one and has a jump there. Changing only the boundary value cannot make two different nearby heights agree.

Example 5: a reciprocal with a vertical asymptote

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

Condition one fails: the original denominator vanishes at three. Condition two also fails: inspect the two signed approaches separately.

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

lim_{x \to 3^{+}} q (x) = \infty

The sides have opposite infinite directions: there is no common finite two-sided limit and no common signed infinite limit. The function is not continuous at three and has an infinite discontinuity associated with a vertical asymptote. A finite target reassignment cannot change that unbounded nearby behavior.

5. A defined point can still have the wrong value

A missing value is not the only removable case. A special point value can miss the finite nearby limit:

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

u (2) = 5, lim_{x \to 2} u (x) = 4

Conditions one and two pass: the value is defined and the finite limit exists. Condition three fails because they differ. Redefining the value to four repairs continuity.

The open point

(2, 4)

gives the limiting height; the filled point

(2, 5)

gives the assigned value. Read them separately.

First locate the filled point at the target, then trace the curve from each available side. See estimating limits from graphs.

6. Connect failed conditions to discontinuity types

The main discontinuity types and a caution
Pattern	Nearby behavior	Assigned value	Can one point repair it?
Removable	A common finite two-sided limit exists.	Missing, or defined and different from the limit.	Yes: assign the finite nearby limit.
Jump	Both side limits are finite but unequal.	May be defined or missing.	No: changing one value cannot change the side limits.
Infinite	At least one relevant side tends to a signed infinity.	May be missing or separately assigned.	No finite reassignment removes unbounded behavior.
Other failure	For example, a side oscillates without a finite or signed infinite limit.	Check separately.	Do not force this into the finite-jump category.

A failed condition alone does not always identify the type. An undefined value is not automatically a hole, and a failed two-sided limit is not automatically a jump. A function with a vertical asymptote can even have an artificially assigned finite point on that line; its finite-limit condition still fails.

Review Topic 1.10: types of discontinuities for more classifications.

7. Standard continuous functions and their domains

Known continuity results make many checks shorter. The qualification is at points in the real domain; transformed expressions still need their own domain check.

Continuous function families: use valid real inputs
Family	Typical expression	Where continuity holds
Polynomial	$2 x^{2} + 3 x - 1$	At every real input.
Rational	$\frac{P (x)}{Q (x)}$	Where the denominator is nonzero.
Exponential	$a^{x}$ or $e^{x}$	At every real input for a positive exponential base.
Logarithmic	$\log_{a} (x)$ or $\ln (x)$	For positive arguments; a logarithm base is positive and different from one.
Trigonometric	$\sin (x)$ , $\cos (x)$ , $\tan (x)$	Throughout the real domain of the particular function.
Root	$\sqrt[n]{x}$	For a positive integer index: odd roots allow all real inputs; even roots require nonnegative radicands.

Operations and compositions

Sums, differences, products and constant multiples of functions continuous at a point are continuous there. A quotient is continuous when the component functions are continuous and the denominator is nonzero at the target. A composition is continuous when the inner function is continuous at the input and the outer function is continuous at the resulting inner value.

lim_{x \to c} f (g (x)) = f (g (c))

Trigonometric inputs are measured in radians in calculus. Do not infer continuity of an arbitrary unknown function merely because substitution works in one part of its expression. Cite an applicable continuity result and verify its conditions.

8. Read continuity from graphs and numerical data

Assigned value: read the height of the included point. An open circle alone does not assign a value.
Left approach: follow the graph from smaller inputs without jumping onto a point-only branch.
Right approach: follow it from larger inputs and compare the finite approached heights.
Final comparison: the common finite height must equal the included point's height. At a domain endpoint, use the relevant one-sided convention.

A corner or vertical tangent does not by itself prove discontinuity. A dashed vertical line does not prove an asymptote: the unbounded limit behavior matters. Graph resolution and rounding can hide exceptional points, so inspect domain restrictions and exact labels.

A finite table can suggest matching approaches without establishing a limit or determining the assigned value. Without additional assumptions, sampled values alone cannot certify continuity. Connect formula, graph and numerical evidence in Topic 1.9: multiple representations.

Separate the value, the limit and their equality
Question	Evidence to find	What it establishes
Is the point included?	$f (c)$ is a real number.	Condition one only.
Do nearby values settle?	$lim_{x \to c} f (x)$ is finite, with matching approaches.	Condition two only.
Do they match?	$lim_{x \to c} f (x) = f (c)$	Condition three, once both exist.
Are all three true?	Defined value, finite common limit and equality.	Continuity at the target.

9. Endpoints: use the approach in the domain

The two-sided test is the usual interior-point test. At the left endpoint of an interval, continuity uses the right-hand approach; at the right endpoint, it uses the left-hand approach. A missing approach outside the domain is not an automatic discontinuity.

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

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

For example, the real square-root function is defined at zero and approaches its assigned value from the right:

lim_{x \to 0^{+}} \sqrt{x} = 0 = \sqrt{0}

It is continuous at zero relative to its domain. Keep that domain explicit rather than substituting negative real inputs into the square root. Extend the test in Topic 1.12: continuity over an interval.

10. Choose a value that makes a function continuous

For a parameter used only at the target, find the nearby limit first, then choose the assigned value to match it. A finite common limit is essential.

Extension example: determine the parameter

F (x) = {\begin{cases} \frac{x^{2} - 1}{x - 1} & if x \neq 1 \\ k & if x = 1 \end{cases}

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

lim_{x \to 1} F (x) = 2

F (1) = k = 2

Choosing two makes all three conditions hold. If the finite side limits differed, changing a target-only parameter could not repair continuity. For a parameter inside a branch formula, solve for matching side limits, then check the included value and domain restrictions.

Continue with Topic 1.13: removing discontinuities.

11. Write a clear AP Calculus justification

When asked to use the definition, organize your response around the assigned value, the finite limit and their equality. At a piecewise boundary, show the one-sided calculations supporting the two-sided limit.

State the value: identify the branch or substitution that gives it.
Establish the limit: use matching finite side limits, an applicable continuity theorem or valid algebra.
Compare and conclude: say whether equality holds and name the target input.

For the polynomial example, cite polynomial continuity and support the conclusion with this chain:

lim_{x \to 2} f (x) = 13 = f (2)

A three-bullet response is an organizational tool, not a guarantee of an exam score. The required justification depends on the particular question and its scoring guidelines.

Typical questions

Is the function continuous at a specified input? Justify using the definition.
Find and classify all discontinuities of a formula, graph or piecewise function.
Choose a target value or branch parameter that makes the function continuous.
Use a graph to compare side limits with the included point.

Efficient shortcuts, with reasons

Polynomial: cite continuity at every real input.
Rational: verify the target denominator is nonzero.
Piecewise: check branch boundaries explicitly.
Composition: check continuity of the inner function and of the outer function at the inner output.

12. Common mistakes and their corrections

Checking only the assigned value. A function can be defined at a jump or have a special value missing the nearby limit.
Using a point-only branch for a side limit. Use the neighboring rule followed by inputs approaching from that side.
Confusing the limit and value. Read open and filled markers separately; one can exist without the other.
Combining opposite infinities into one two-sided result. Give each side its own sign. Opposite infinities do not form a common signed limit.
Calling every denominator zero a hole. A finite common limit is needed. Cancellation can leave another zero in the denominator.
Losing the original domain. Simplifying does not fill an excluded input.
Equating continuity with smoothness. A corner can be continuous; differentiability is stronger.
Ignoring endpoints or excluded inputs. Identify the domain, then use the relevant approach. An undefined value prevents continuity of the original function there, even when an extension is possible.
Giving an unsupported conclusion. Show the value, limit evidence and comparison appropriate to the question.

A picture guides the test; it does not replace it

The absence of a visible break on a calculator screen is not a proof, and a sharp bend is not automatically a discontinuity. Compare the finite nearby limit with the actual assigned value.

13. Four practice problems with complete explanations

Find the assigned value, the relevant limit and their comparison before opening each solution. All four original practice problems are retained.

1. A linear function at input two

f (x) = 3 x + 5

Check the continuity test and explanation

f (2) = 11, lim_{x \to 2} f (x) = 11

A linear function is a polynomial, so direct substitution is justified. The value is defined, the finite limit exists and both equal eleven. The function is continuous at two.

2. A rational hole at input three

g (x) = \frac{x^{2} - 9}{x - 3}

Check the continuity test and explanation

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

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

The original function is undefined at three, so condition one fails. The finite limit is six, but there is no assigned value to match it. This is a removable hole at $(3, 6)$ . Assigning six at that input gives a continuous extension.

3. A piecewise boundary at input one

h (x) = {\begin{cases} x^{2} & if x \leq 1 \\ 2 x - 1 & if x > 1 \end{cases}

Check the continuity test and explanation

h (1) = 1

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

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

The first branch includes the target and gives one. Both neighboring rules also approach one. All three conditions hold, so the function is continuous at one.

4. A reciprocal at input four

p (x) = \frac{2}{x - 4}

Check the continuity test and explanation

lim_{x \to 4^{-}} p (x) = - \infty

lim_{x \to 4^{+}} p (x) = \infty

The original value at four is undefined and the side limits give no finite two-sided limit. The function is not continuous there; it has an infinite discontinuity and vertical asymptote. No finite target reassignment repairs it.

14. Quick reference: defined, finite, equal

The continuity test at a point
Check	Required evidence	If it fails
Assigned value	$f (c)$ is a real number.	No continuity as originally defined; still inspect the limit if classification or repair is requested.
Nearby limit	$lim_{x \to c} f (x)$ is finite. At an interior point, both approaches agree.	No finite two-sided continuity there. Determine the side behavior before classifying.
Equality	$lim_{x \to c} f (x) = f (c)$ .	If both quantities exist but differ, the discontinuity is removable.
All three pass	Defined value, finite relevant limit and equality.	The function is continuous at the target.

At a domain endpoint, replace the two-sided limit with the relevant one-sided limit. A clear final statement names the input and gives the reason for continuity or its failure.

15. Why continuity matters and what to study next

Continuity connects limits with later calculus. Verify it before applying a theorem that requires it instead of assuming it from a plausible graph.

Intervals: Topic 1.12 checks each relevant point, with appropriate endpoint conventions.
Repair: Topic 1.13 uses finite limiting values to remove holes or select parameters.
Infinite limits: Topic 1.14 links signed unbounded behavior with vertical asymptotes.
Intermediate Value Theorem: Topic 1.16 needs continuity on a closed interval to guarantee intermediate output values.
Derivatives: differentiability at an interior point implies continuity, but continuity alone does not guarantee a derivative.
Integrals: continuity on a closed interval is a useful sufficient condition for integrability. Some discontinuous functions are also integrable, so continuity is not a universal requirement for every definite integral.

Previous: Topic 1.10 — Discontinuity types →Next: Topic 1.12 — Continuity over an interval →Topic 1.13 — Removing discontinuities →All Unit 1 study notes →

After reviewing the remaining unit topics, apply the definitions in the Unit 1 practice test collection. Keep the three tasks separate: find the value, establish the finite limit and compare them.

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