Names the property that makes substitution valid and states it as three conditions. Classifies discontinuities by which condition fails, distinguishes removable failures from the rest, and states the intermediate value property that continuity guarantees.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 12 — Introduction to Calculus
§12.3 Continuity, pp. 1416-1431
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §12.3 Continuity §12.3, pp. 1416-1431 — the pages these objectives are drawn from
Warm-up
The previous section used it constantly without saying why.
Discussion prompt
Substituting the point gives the limit for most functions at most points. What property is that?
Hint: What has to be true of the function there?
Answer:
The value and the limit have to agree — what the function is at the point matches what it approaches nearby.
And both have to exist in the first place. That is three requirements: a value, a limit, and their agreement.
A function satisfying all three at a point is continuous there. Substitution works exactly where continuity holds, which is why it works so often and fails so informatively.
Concept
A function is continuous at a point when it has a value there, has a limit there, and the two are equal. Failing any one produces a discontinuity.
continuous — having a value and a limit at a point which are equal, so that the graph has no break there
\[ \lim_{x\to a}f(x)=f(a) \]
Writing it as one equation hides three claims: that the limit exists, that the value exists, and that they match. Taking them separately is what makes the classification of discontinuities systematic.
Figure (svg): Three cards giving the three conditions a function must satisfy to be continuous at a point
OpenStax, Precalculus, §12.3 Continuity §12.3, pp. 1416-1420
Section
Section 1
Concept
Continuity at a point requires three separate things, and checking them in order identifies exactly which one fails when the function is discontinuous.
The order matters for diagnosis rather than for the definition. Checking the limit first is often quicker, since its failure rules out continuity regardless of what the value does.
Figure (svg): Three cards giving the three conditions a function must satisfy to be continuous at a point
OpenStax, Precalculus, §12.3 Continuity §12.3, pp. 1416-1421
Picture it
Each card names one and how it fails.
Figure (svg): Three cards giving the three conditions a function must satisfy to be continuous at a point
The notes on each card are the classification in miniature. Which condition fails determines which kind of discontinuity the function has.
Worked example
Three checks in order.
\[ \text{Is } f(x)=\frac{x^2-4}{x-2} \text{ continuous at } x=2? \]
Check the value
Why: Substituting gives zero over zero.
Note the first condition fails
Why: No value exists.
Check the limit anyway
Why: By factoring.
\[ 4 \]
Classify
Why: Limit exists, value does not.
Figure (svg): Three cards giving the three conditions a function must satisfy to be continuous at a point
\[ \text{removable discontinuity at }2 \]
Verify: check what redefining would do
Why: Setting the value at 2 to be 4 would make all three conditions hold, so one redefinition fixes it. That possibility is exactly what removable means, and it depends on the limit existing.
OpenStax, Precalculus, §12.3 Continuity §12.3, pp. 1417-1419
Sorting
Three conditions, three failures.
Sort into buckets
Sort each situation.
Worked example
The ordinary case.
\[ \text{Is } f(x)=x^2+1 \text{ continuous at } x=3? \]
Check the value
Why: Substituting.
\[ 10 \]
Check the limit
Why: By the properties.
\[ 10 \]
Compare
Why: They agree.
Conclude
Why: All three hold.
Figure (svg): The solution to Worked example all three hold shown as a ladder of expressions, one row per legal move
\[ \text{continuous} \]
Verify: note where this holds
Why: A polynomial satisfies all three conditions at every point, which is why substitution always works for one. That is a general fact about polynomials rather than a feature of this particular point.
OpenStax, Precalculus, §12.3 Continuity §12.3, pp. 1419-1421
Trap
\[ f(2)\text{ is defined, so }f\text{ is continuous at }2 \]
Take a defined value as sufficient
Why: The limit and the agreement are not checked.
A jump discontinuity can have a perfectly well defined value at the point.
All three conditions must hold. A value alone is not enough.
A function can have a value at a point where the limit does not exist, or where it exists and disagrees.
Checking the limit is usually the quicker route, since its failure rules out continuity without needing the value at all.
Prediction
A function has a value at the point.
Predict first
Is it continuous there?
Correct: Not necessarily; the limit must exist and agree.
Why: A jump discontinuity can have a perfectly well defined value at the point while the limit fails to exist. All three conditions are needed, and a value alone is the weakest of them.
Faded example
As a single equation.
Fill in the blanks
\lim_af(x)=f(3), \text______\text___
Why: The equation asserts that the limit exists, that the value exists, and that they are equal. Taking them separately is what makes the classification of failures systematic.
Step zero
You are testing continuity at a point.
Discussion prompt
Which condition do you check first, and why?
Hint: Which failure is most decisive?
Answer:
The limit, because its failure rules out continuity immediately and without needing the value at all.
If the limit exists, then check the value and their agreement — both of which are quick once the limit is known.
Checking in this order also classifies the failure as it goes: no limit means a jump or infinite discontinuity, and a limit without agreement means a removable one.
Section
Section 2
Concept
A discontinuity is removable if the limit exists, a jump if the two sides disagree, and infinite if the outputs grow without bound.
The names are descriptive and the classification is complete for the functions met at this level. Each kind corresponds to one way the three conditions can fail, which is why there are three.
| kind | what fails | what it looks like |
|---|---|---|
| removable | the value, or its agreement | a hole in a smooth curve |
| jump | the limit, by disagreement | a break with two pieces |
| infinite | the limit, by unbounded growth | a vertical asymptote |
Figure (svg): Three graphs showing a removable discontinuity, a jump discontinuity and an infinite discontinuity
OpenStax, Precalculus, §12.3 Continuity §12.3, pp. 1421-1425
Picture it
A hole, a break and an asymptote.
Figure (svg): Three graphs showing a removable discontinuity, a jump discontinuity and an infinite discontinuity
The bottom line anticipates the next idea. Only the first can be repaired, and the reason is that only there does a limit exist to repair it to.
Worked example
Find which condition fails.
\[ \text{Classify the discontinuity of } \frac{1}{x-3} \text{ at } x=3. \]
Check the value
Why: Division by zero.
Check the limit
Why: Nonzero over zero.
Identify the behaviour
Why: The outputs run away.
Classify
Why: By the behaviour.
Figure (svg): Three graphs showing a removable discontinuity, a jump discontinuity and an infinite discontinuity
\[ \text{infinite} \]
Verify: check the graph's shape
Why: The curve rises without bound on one side of 3 and falls without bound on the other, which is a vertical asymptote. No single value could be assigned there to repair it.
OpenStax, Precalculus, §12.3 Continuity §12.3, pp. 1422-1424
Sorting
The behaviour names it.
Sort into buckets
Sort each description.
Worked example
The two sides disagree.
\[ \text{Classify the discontinuity of a function that is } 1 \text{ below } 0 \text{ and } 2 \text{ at or above.} \]
Check the value
Why: Defined, equal to two.
Check the one-sided limits
Why: One from below, two from above.
Note the limit fails
Why: By disagreement.
Classify
Why: Two pieces at different heights.
Figure (svg): The solution to Worked example a jump shown as a ladder of expressions, one row per legal move
\[ \text{jump} \]
Verify: note that the value existed
Why: The function is perfectly well defined at zero, so the first condition held — it was the limit that failed. That shows a value alone is not enough, which is the trap from the previous idea.
OpenStax, Precalculus, §12.3 Continuity §12.3, pp. 1424-1425
Error analysis
A student classifies a discontinuity where the graph breaks.
Annotate
On: \( \text{the graph breaks at }0, \text{ so redefining }f(0)\text{ will fix it} \)
Removability is entirely about whether a limit exists. A jump has two well-defined one-sided limits and no two-sided one, so there is nothing for a single value to match.
Prediction
Substitution gives a nonzero number over zero.
Predict first
What kind of discontinuity is it?
Correct: Infinite.
Why: A nonzero numerator over a vanishing denominator means the outputs grow without bound, which is a vertical asymptote. The limit fails, so the discontinuity is not removable.
Faded example
One-sided limits of one and two.
Fill in the blanks
\lim_1f=2 \ne ___=\lim____f
Why: Both one-sided limits exist but disagree, which is exactly what a jump is. The two-sided limit failing is what makes it non-removable.
Explain it
Each name describes what the graph does.
Discussion prompt
Explain what each kind looks like and why it is called that.
Hint: Match each to a picture.
Answer:
Removable is a hole in an otherwise smooth curve, and it is called that because filling the hole repairs it.
Jump is a break where the curve continues at a different height, which is what jumping describes.
Infinite is a vertical asymptote, where the outputs grow without bound. A good explanation notes that all three names are descriptive, so the classification can be read off a graph without any computation.
Section
Section 3
Concept
Redefining a function at one point can only fix a discontinuity if the limit exists there, since the new value must equal the limit.
This makes removability a property of the limit rather than of the value. Whether the value is missing or merely wrong makes no difference — what matters is whether a limit exists for it to be set to.
Figure (svg): A contrast between a removable discontinuity, which one redefinition fixes, and the others, which no redefinition can
OpenStax, Precalculus, §12.3 Continuity §12.3, pp. 1425-1428
Picture it
The first row on each side is the deciding fact.
Figure (svg): A contrast between a removable discontinuity, which one redefinition fixes, and the others, which no redefinition can
The caption states the criterion. Everything else follows from whether a limit exists, which is why the classification and the removability question have the same answer.
Worked example
Set the value to the limit.
\[ \text{How would you make } \frac{x^2-4}{x-2} \text{ continuous at } 2? \]
Find the limit
Why: By factoring.
\[ 4 \]
Note the value
Why: Undefined.
Define it to match
Why: Set the value at 2 to 4.
\[ f(2) = 4 \]
Check all three conditions
Why: Now all hold.
Figure (svg): A contrast between a removable discontinuity, which one redefinition fixes, and the others, which no redefinition can
\[ f(2)=4 \]
Verify: check the redefinition works
Why: With that value, the limit is 4 and the value is 4, so they agree — all three conditions hold. The redefinition changed the function at exactly one point and repaired it entirely.
OpenStax, Precalculus, §12.3 Continuity §12.3, pp. 1426-1427
Prediction
A discontinuity where the limit exists but the value is wrong.
Predict first
Is it removable?
Correct: Yes, by setting the value to the limit.
Why: Removability depends on the limit existing, not on whether the value is missing or wrong. Either way, redefining the single value to match the limit achieves continuity.
Worked example
Nothing to set the value to.
\[ \text{Why can redefining not fix a jump discontinuity?} \]
Note what redefining does
Why: Changes one value.
Note what continuity needs
Why: The value to equal the limit.
Note the jump's limits
Why: Two different one-sided values.
Conclude
Why: No single value matches both.
Figure (svg): The solution to Worked example why a jump cannot be removed shown as a ladder of expressions, one row per legal move
\[ \text{no target for the value} \]
Verify: consider trying each one-sided value
Why: Setting the value to the left-hand limit still leaves the right side disagreeing, and vice versa. Neither choice achieves continuity, which is what having no two-sided limit means.
OpenStax, Precalculus, §12.3 Continuity §12.3, pp. 1427-1428
Trap
\[ \text{the function is undefined there, so redefining will fix it} \]
Take the missing value as the whole problem
Why: The limit is not checked.
A function can be undefined at a point where the limit also fails, as at an asymptote.
The limit is what decides removability, not the value. A missing value is fixable only if there is a limit to match it to.
At an asymptote the value is missing and the limit fails too, so no redefinition helps.
Check the limit first. If it exists the discontinuity is removable, whether the value is missing or merely wrong.
Sorting
The limit decides.
Sort into buckets
Sort each situation.
Faded example
With a limit of four at the point.
Fill in the blanks
\text4f(2)=agree, \text______
Why: Setting the value to the limit makes all three conditions hold at once. That single redefinition is what removable means, and it is available only when the limit exists.
Explain it to yourself
Removability has one condition.
Discussion prompt
Explain what it is and why.
Hint: What must the new value equal?
Answer:
Redefining changes the function at one point, and continuity requires that value to equal the limit.
So a limit must exist for there to be anything to set the value to. If it does not — a jump or an asymptote — no choice of value works.
Whether the value is currently missing or wrong is irrelevant, since either way it is being replaced. A good explanation notes that removability is therefore a property of the limit alone, which is why checking the limit first classifies and decides at once.
Section
Section 4
Concept
A function is continuous on an interval when it is continuous at every point of it, which for the standard functions means avoiding the points where they misbehave.
The combination rules mirror the limit properties exactly, and for the same reason: continuity is defined by a limit condition, so anything the limit properties permit, continuity inherits.
Figure (svg): Three graphs showing a removable discontinuity, a jump discontinuity and an infinite discontinuity
OpenStax, Precalculus, §12.3 Continuity §12.3, pp. 1428-1430
Picture it
Three kinds of point to avoid.
Figure (svg): Three graphs showing a removable discontinuity, a jump discontinuity and an infinite discontinuity
An interval avoiding all such points is one on which the function is continuous throughout. For the standard functions those points are easy to identify in advance.
Worked example
Identify the bad points.
\[ \text{Where is } \frac{x+1}{x^2-9} \text{ continuous?} \]
Identify the function type
Why: Rational.
Factor the denominator
Why: A difference of squares.
\[ (x - 3) (x + 3) \]
Find its zeros
Why: Two points.
\[ 3\text{ and } -3 \]
State the answer
Why: Everywhere else.
Figure (svg): The solution to Worked example find where a function is continuous shown as a ladder of expressions, one row per legal move
\[ \text{all }x\ne\pm 3 \]
Verify: check what happens at the bad points
Why: At both points the numerator is nonzero while the denominator vanishes, giving infinite discontinuities. Neither is removable, so the exclusions are permanent rather than repairable.
OpenStax, Precalculus, §12.3 Continuity §12.3, pp. 1428-1429
Sorting
Each function type has its own trouble spots.
Sort into buckets
Sort each function type.
Worked example
The domain restricts the interval.
\[ \text{Where is } \sqrt{x-2} \text{ continuous?} \]
Find the domain
Why: The radicand must be non-negative.
\[ x\text{ at least } 2 \]
Note the function type
Why: A root, continuous on its domain.
Consider the endpoint
Why: Only one side exists.
State the answer
Why: The whole domain.
\[ x\text{ at least } 2 \]
Figure (svg): The solution to Worked example a root function shown as a ladder of expressions, one row per legal move
\[ x\ge 2 \]
Verify: note the endpoint's status
Why: At 2 the function is continuous from the right only, since there is nothing to its left. Endpoints of a domain always have this one-sided character, which is a technicality rather than a discontinuity.
OpenStax, Precalculus, §12.3 Continuity §12.3, pp. 1429-1430
Error analysis
A student describes where a function is continuous.
Annotate
On: \( \frac{x^2-4}{x-2}: \text{ discontinuous at }2, \text{ so it can never be continuous there} \)
Whether an exclusion is permanent depends on the kind of discontinuity. A removable one can be repaired by redefining, and the classification is what says which case applies.
Prediction
A rational function with denominator x squared minus nine.
Predict first
Where is it discontinuous?
Correct: At three and negative three.
Why: The denominator vanishes at those two points, and division by zero makes the function undefined there. Everywhere else the quotient property applies and continuity holds.
Faded example
For a denominator factoring as a difference of squares.
Fill in the blanks
x^2-9=(x-3)(x+3)=0 \;\Longrightarrow\; x=\pm3
Why: Setting the denominator to zero locates the points where the function is undefined. Those are the only candidates for discontinuity in a rational function.
Explain it
Sums and products of continuous functions are continuous.
Discussion prompt
Explain to a classmate why that follows.
Hint: How is continuity defined?
Answer:
Continuity is defined by a limit condition — the limit equalling the value.
The limit properties say a limit passes through sums, products and quotients, so if the parts satisfy the condition then so does the combination.
The quotient carries the same restriction it did for limits, needing a nonzero denominator. A good explanation notes that the combination rules are the limit properties restated, rather than a separate set of facts.
Section
Section 5
Concept
If a continuous function is below a level at one end of an interval and above it at the other, it must equal that level somewhere between.
The property guarantees existence without locating the point, which is exactly what root-finding methods need — they establish that a solution is in an interval and then narrow it down.
Figure (svg): A continuous curve crossing a horizontal level between two endpoints, illustrating the intermediate value property
OpenStax, Precalculus, §12.3 Continuity §12.3, pp. 1430-1431
Picture it
Below at one end, above at the other.
Figure (svg): A continuous curve crossing a horizontal level between two endpoints, illustrating the intermediate value property
The bottom line is the reason it holds. A continuous curve has no breaks to jump the level with, so it must cross somewhere.
Worked example
Opposite signs at the ends.
\[ \text{Show that } x^3+x-3 \text{ has a root between } 1 \text{ and } 2. \]
Check continuity
Why: A polynomial.
Evaluate at the left end
Why: One plus one minus three.
\[ -1 \]
Evaluate at the right end
Why: Eight plus two minus three.
\[ 7 \]
Apply the property
Why: Opposite signs.
Figure (svg): A continuous curve crossing a horizontal level between two endpoints, illustrating the intermediate value property
\[ \text{a root in }(1,2) \]
Verify: note what has and has not been shown
Why: The existence of a root is established without finding it. The property gives no location beyond the interval, which is why numerical methods follow it up by halving the interval repeatedly.
OpenStax, Precalculus, §12.3 Continuity §12.3, pp. 1430-1431
Prediction
A continuous function is negative at one end and positive at the other.
Predict first
What follows?
Correct: It equals zero somewhere between.
Why: The property guarantees a crossing without locating it or counting how many there are. It is an existence statement, which is exactly what root-finding methods need as a starting point.
Worked example
A discontinuous function can skip.
\[ \text{Why does the property need continuity?} \]
Consider a jump
Why: The graph breaks.
Place the level in the gap
Why: Between the two heights.
Check the endpoints
Why: Below at one, above at the other.
Conclude
Why: The level is skipped.
Figure (svg): The solution to Worked example why continuity is required shown as a ladder of expressions, one row per legal move
\[ \text{continuity prevents skipping} \]
Verify: identify what continuity provides
Why: Continuity means no breaks, so the curve must pass through every intermediate height rather than jumping past. That unbroken passage is exactly what the property depends on.
OpenStax, Precalculus, §12.3 Continuity §12.3, pp. 1431-1431
Trap
\[ \text{the property gives a root, so it tells us where it is} \]
Read existence as location
Why: The property is taken to identify the point.
It guarantees a crossing somewhere in the interval and says nothing more.
The property is an existence statement. It establishes that a crossing occurs without identifying where.
That is still valuable: knowing a root is in an interval is what allows numerical methods to narrow it down.
Existence and location are different questions, and this property answers only the first.
Sorting
It needs continuity and opposite sides.
Sort into buckets
Sort each situation.
Faded example
Evaluating a polynomial at two endpoints.
Fill in the blanks
f(1)=-1\text2f(2)=7: \text___1\text______
Why: Opposite signs at the endpoints of an interval where the function is continuous forces a crossing of zero somewhere inside. The property gives no more than that.
Explain it to yourself
The property fails without it.
Discussion prompt
Explain what continuity provides.
Hint: What could a broken curve do?
Answer:
A jump lets the curve move from below the level to above it without ever taking it — the level falls in the gap.
Continuity forbids that, since it means no breaks: the curve must pass through every intermediate height on its way.
So the property is a direct consequence of there being no gaps. A good explanation notes that this is the practical payoff of continuity — it converts a statement about smoothness into a guarantee that solutions exist.
Comparison
Fill the blanks from memory. Which condition fails names the kind.
Comparison matrix
| removable | jump | infinite | |
|---|---|---|---|
| does the limit exist | yes | no, the sides disagree | no, unbounded |
| looks like | a hole | a break | an asymptote |
| removable | yes | no | no |
| why | a limit to match | nothing to match | nothing to match |
The first row decides the third. Removability depends entirely on whether a limit exists for the value to be set to.
Pattern
Five steps, and the first classifies as it goes.
Checking the limit first both decides continuity and classifies any failure, where checking the value first settles neither on its own.
Check
The definition.
Check your understanding
How many conditions must hold for continuity at a point?
Answer: A
Why: The single equation stating that the limit equals the value asserts all three: that the limit exists, that the value exists, and that they match. Each can fail separately, which is what the classification catalogues.
Check
Removability.
Check your understanding
What makes a discontinuity removable?
Answer: A
Why: Redefining sets one value, and continuity requires that value to equal the limit. So a limit must exist for there to be anything to set it to, regardless of whether the current value is missing or wrong.
Check
The intermediate value property.
Check your understanding
What does it guarantee?
Answer: A
Why: It is an existence statement: a continuous curve straddling a level must cross it. It says nothing about where or how many times, which is why numerical methods follow it up.
Real world
Numerical root-finding is built on the intermediate value property.
Discussion prompt
A computer finds a root by repeatedly halving an interval. Why does that work?
Hint: What is known at each step?
Answer:
The method starts with an interval where the function has opposite signs at the ends, so the property guarantees a root inside.
Halving the interval and checking the sign at the midpoint identifies which half still straddles zero — and the guarantee carries over to that half.
Repeating narrows the interval as tightly as required, with the property assuring at every stage that a root is still inside. The existence guarantee is what makes the search terminate meaningfully, rather than merely producing a number that might be nothing.
Commit first
State your confidence along with your answer.
Predict first
Why can a jump discontinuity not be removed?
Correct: There is no two-sided limit for the value to equal.
Why: Redefining changes one value, and continuity requires it to equal the limit. A jump has two disagreeing one-sided limits and no two-sided one, so no single value could satisfy both sides.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
A classmate says a function is continuous because it has a value at the point. Explain what is missing.
Hint: How many conditions are there?
Answer:
A value is only the first of three conditions. The limit must also exist, and the two must agree.
A jump discontinuity has a perfectly good value at the point while the limit fails entirely, so a value alone proves nothing.
Checking the limit first is usually quicker, since its failure settles the question without needing the value. A good explanation notes that this order also classifies the failure as it goes, which the value-first order does not.
Exit ticket
One honest answer, so the next lesson can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The third is where the classification becomes reasoning rather than recall, and the fourth is the first result in the course that guarantees something exists without finding it.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Write the three conditions and note which kind of discontinuity each failure produces. Beside them, sketch all three kinds and mark which is removable. Underneath, explain in one sentence why removability depends on the limit, and sketch a curve crossing a level to illustrate the intermediate value property.
If your removability sentence refers to the limit rather than the value, the section's central reasoning is on the page rather than its classification alone.
Recap
Five things, and the first organises the rest.
| if you remember one thing | it should be this |
|---|---|
| about the definition | three conditions, and each fails differently |
| about testing | check the limit first; it decides and classifies at once |
| about removability | it depends on the limit existing, never on the value |
| about the property | it guarantees existence and says nothing about location |
Section 12.4 closes the course with the derivative, where a limit of the kind resolved in §12.2 turns an average rate of change into an instantaneous one.
OpenStax, Precalculus, §12.3 Continuity §12.3, pp. 1416-1431 — everything on these slides traces back here
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