The three-part definition of continuity at a point and the three ways it fails, the classification of discontinuities as removable, jump or infinite, continuity on an interval with one-sided continuity at endpoints, the algebra of continuous functions and continuity of composites, and the Intermediate Value Theorem as the course's first existence theorem.
Subject: Calculus I · 65 slides · symbolic lesson
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Title
Calculus I · Chapter 2 — Limits
Continuity
Objectives
Five outcomes. The last is the first theorem in this course that guarantees something exists without telling you what it is.
OpenStax Calculus Volume 1, §2.4 Continuity §2.4, pp. 155-168 — the section these objectives are drawn from
Warm-up
Section 2.3 established that for a polynomial the limit at a point equals the value there. That is what made substitution legitimate.
Discussion prompt
For which functions does substitution give the limit, and for which does it not? What exactly is the property that separates them?
Hint: Compare the limit and the value at the point, and ask when they agree.
Answer:
\[ \lim_{x \to a} f(x) = f(a) \quad \text{- when does this hold?} \]
For every polynomial it holds; for a rational function it holds except where the denominator vanishes; for a piecewise function with a jump it fails at the seam.
That equation IS the property, and this section gives it a name: continuity. Once named, it stops being a convenient coincidence and becomes a hypothesis that theorems can be built on — including the first existence theorem in the course.
Concept
A function is continuous at a point when three things hold: the function is defined there, the limit exists there, and the two are equal. Informally, the graph can be drawn through that point without lifting the pen.
continuity at a point — A function is continuous at a if f of a is defined, the limit of f at a exists, and the limit equals f of a. Failure of any one of the three makes the function discontinuous there.
\[ f \text{ continuous at } a \iff \lim_{x \to a}f(x) = f(a) \]
The single equation packs all three conditions, since writing it presupposes both sides exist. Separating them is worth doing anyway, because the three failures behave very differently and only one of them can be repaired.
Figure (svg): The three conditions for continuity, each shown failing in its own way
OpenStax Calculus Volume 1, §2.4 Continuity §2.4, pp. 155-158
Section
Section 1
Concept
Continuity at a point requires all three. The value must exist, the limit must exist, and they must be the same number. Each can fail while the other two hold, which is why the definition is stated in three parts.
the three conditions — First, f of a is defined. Second, the limit of f as x approaches a exists. Third, that limit equals f of a. All three are required, and each can fail independently of the others.
\[ (1)\; f(a) \text{ exists} \quad (2)\; \lim_{x \to a}f(x) \text{ exists} \quad (3)\; \text{they are equal} \]
Section 2.2 spent its time insisting that the limit and the value are independent quantities computed from disjoint sets of inputs. Continuity is precisely the extra demand that, despite that independence, they coincide.
Figure (svg): The three conditions for continuity, each shown failing in its own way
OpenStax Calculus Volume 1, §2.4 Continuity §2.4, pp. 155-159 — continuity at a point
Picture it
One picture for each way of breaking continuity.
Figure (svg): The three conditions for continuity, each shown failing in its own way
The first has a hole, the second a jump, the third a misplaced value. Each satisfies two of the three conditions and fails the remaining one, which is why all three clauses are needed.
Worked example
Example 2.26. Check them in order and stop at the first failure.
\[ \text{Is } f(x) = \frac{x^2-4}{x-2} \text{ continuous at } x = 2? \]
Check whether the value exists
Why: The denominator vanishes at 2.
\[ f(2)\text{ is undefined} \]
Stop: condition one already fails
Why: No further checking can rescue it.
\[ \text{not continuous at } 2 \]
Check the limit anyway, to classify the break
Why: Factor and cancel.
\[ \lim = 4 \]
Note what kind of failure this is
Why: The limit exists but the value does not.
Figure (svg): The solution to Worked example testing all three conditions shown as a ladder of expressions, one row per legal move
\[ f(2) \text{ undefined}, \quad \lim_{x \to 2}f(x) = 4 \]
Verify: confirm it really can be repaired
Why: Defining f of 2 to be 4 makes all three conditions hold, and the repaired function is x plus 2 everywhere — perfectly continuous. That the break can be removed by changing a single value is exactly what removable means. Note the checking order saved work: once condition one failed, continuity was settled, and the limit was computed only to classify the failure rather than to decide it.
OpenStax Calculus Volume 1, §2.4 Continuity §2.4, pp. 157-158
Sorting
Test the three in order at the point named.
Sort into buckets
Sort each function.
The reciprocal at 0 fails condition one, and also condition two — it is unbounded there. When several conditions fail at once, the classification is decided by the one-sided limits, which is the next idea.
Worked example
Checkpoint 2.26. All the pieces are present and still it fails.
\[ \text{Is } f(x) = \begin{cases} \frac{x^2-4}{x-2} & x \ne 2 \\ 3 & x = 2 \end{cases} \text{ continuous at } 2? \]
Check the value
Why: The second piece supplies it.
\[ f(2) = 3,\text{ defined} \]
Check the limit
Why: The first piece governs every nearby input.
\[ \lim = 4,\text{ exists} \]
Compare them
Why: Three against 4.
Conclude
Why: Condition three fails.
\[ \text{not continuous at } 2 \]
Figure (svg): The solution to Worked example a value that exists but does not match shown as a ladder of expressions, one row per legal move
\[ \lim_{x \to 2}f(x) = 4 \ne 3 = f(2) \]
Verify: identify which repair would work
Why: Changing f of 2 from 3 to 4 makes it continuous, so this is again a removable discontinuity — despite the function being perfectly well defined at the point. Removable does not mean the value is missing; it means the limit exists, so SOME single value would work. Here the wrong one was chosen, which is a different failure from the previous example and the same classification.
OpenStax Calculus Volume 1, §2.4 Continuity §2.4, pp. 158-159
Trap
\[ f(2) = 3 \text{ is perfectly well defined} \]
Conclude the function is continuous there
Why: The student checks only the first condition.
\[ \text{so } f \text{ is continuous at } 2 \quad \text{(wrong)} \]
The limit is 4 and the value is 3. The graph has a point floating off the curve, and the pen must jump to reach it.
\[ \lim_{x \to 2}f = 4 \ne 3 = f(2) \;\Longrightarrow\; \text{discontinuous} \]
Check all three conditions, in order, every time
Why: Being defined is necessary and nowhere near sufficient.
The converse error is equally common: assuming that an undefined value means no limit. Section 2.2 was built around the fact that a limit can exist where the function does not, and every derivative in Chapter 3 relies on it. Definedness and having a limit are independent; continuity is the demand that both hold AND agree.
Fill the middle
The single equation that packs all three requirements.
Fill in the blanks
f \textf(a) a \iff \lim____ f(x) = ___
Why: Writing the equation presupposes that both sides exist, so it contains all three conditions at once. Separating them into three clauses is still worth doing, because the three failures are classified differently and only one can be repaired.
Two truths and a lie
All three are about the definition.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. Being defined is only the first of three conditions. The piecewise quotient with the value 3 at the input 2 is defined, has the limit 4, and is discontinuous — the graph has a point sitting off the curve. Definedness and continuity are genuinely different properties, and conflating them is the most common error in this section.
Prediction
Commit before reasoning.
Predict first
Why is the definition stated in three parts rather than as the single equation?
Correct: Because there are exactly three ways to fail, and separating them lets you classify the break.
\[ \text{clause 2 holds} \;\Longrightarrow\; \text{removable}; \qquad \text{clause 2 fails} \;\Longrightarrow\; \text{permanent} \]
Why: The single equation is equivalent, since writing it presupposes both sides exist. But when it fails, the three-part version tells you WHICH clause broke — and that determines the classification and whether the break can be repaired. A failure of the first or third clause with the limit intact is removable; a failure of the second is not. That practical difference is what earns the three-part statement its place.
Section
Section 2
Concept
A discontinuity is removable if the limit exists, a jump if both one-sided limits exist but differ, and infinite if at least one of them is unbounded. Only the removable kind can be repaired, and only by changing one value.
removable, jump and infinite discontinuities — A discontinuity is removable when the two-sided limit exists; a jump when both one-sided limits exist and differ; and infinite when at least one one-sided limit is unbounded.
\[ \text{removable} \;|\; \text{jump} \;|\; \text{infinite} \]
The classification is decided entirely by the one-sided limits and never by the value at the point. That is why the removable case includes both a missing value and a misplaced one — the value's behaviour is irrelevant to the classification.
Figure (svg): The three kinds of discontinuity: removable, jump and infinite
OpenStax Calculus Volume 1, §2.4 Continuity §2.4, pp. 159-162 — types of discontinuities
Picture it
A hole, a step, and an asymptote.
Figure (svg): The three kinds of discontinuity: removable, jump and infinite
Only the first can be fixed. Redefining a single value closes a hole, but no single value can bridge a jump or tame an asymptote — the failure there is in the approach itself.
Worked example
Example 2.28. Compute the one-sided limits and read off the kind.
\[ \text{Classify the discontinuity at the point named for } \frac{x^2-4}{x-2}, \; \frac{|x-1|}{x-1}, \; \frac{1}{(x-3)^2}. \]
For the first, compute the limit
Why: Cancel the shared factor.
\[ \lim = 4,\text{ exists} \]
Classify it
Why: The two-sided limit exists.
For the second, take each side
Why: The absolute value flips sign below 1.
\[ \text{left } -1,\text{ right } 1 \]
Classify it
Why: Both exist and differ.
For the third, examine the behaviour
Why: A square keeps the denominator positive and tiny.
Classify it
Why: At least one side is unbounded.
Figure (svg): The solution to Worked example classifying three breaks shown as a ladder of expressions, one row per legal move
\[ \text{removable}, \quad \text{jump}, \quad \text{infinite} \]
Verify: ask which of the three can be repaired
Why: Only the first. Setting f of 2 equal to 4 makes it continuous. For the second, any value chosen at 1 leaves one side disagreeing with it, since the two sides disagree with each other. For the third, no finite value can be approached at all. That exactly one of the three is fixable is the whole reason the classification is worth making.
OpenStax Calculus Volume 1, §2.4 Continuity §2.4, pp. 160-161
Sorting
Compute both one-sided limits first.
Sort into buckets
Sort each discontinuity.
Tangent at pi over 2 is infinite in the sharpest way: one side goes to positive infinity and the other to negative infinity. Both behaviours are unbounded, which is all the classification requires — it does not demand that the two sides agree.
Worked example
Checkpoint 2.28. Find the one value that works.
\[ \text{Define } f(3) \text{ so that } f(x) = \frac{x^2-9}{x-3} \text{ becomes continuous at } 3. \]
Compute the limit at the point
Why: Factor and cancel.
\[ (x - 3) (x + 3) / (x - 3) = x + 3 \]
Evaluate the simplified form
Why: Three plus 3.
\[ \lim = 6 \]
Set the value equal to the limit
Why: This is condition three.
\[ \text{define } f(3) = 6 \]
Confirm all three conditions now hold
Why: Defined, limit exists, and they agree.
Figure (svg): The solution to Worked example repairing a removable discontinuity shown as a ladder of expressions, one row per legal move
\[ f(3) := 6 \;\Longrightarrow\; f \text{ continuous at } 3 \]
Verify: check that no other value works
Why: Any other choice leaves the limit at 6 and the value elsewhere, so condition three fails again. The repair value is forced to be the limit, which is why removable discontinuities have exactly one repair and why the limit is the thing that determines it. Notice too that this is the same manoeuvre as cancelling in Section 2.3 — the simplified expression IS the repaired function.
OpenStax Calculus Volume 1, §2.4 Continuity §2.4, pp. 161-162
Error analysis
A student classifies a discontinuity.
Annotate
On: \( \frac{|x-1|}{x-1}: \; \text{undefined at } 1, \text{ so define } f(1) = 0 \text{ to repair it} \)
A missing value looks like a hole and may not be one. Always compute both one-sided limits before deciding a break is removable — the classification depends on those, never on whether the point has a value.
Fill the middle
Repairing the quotient from the worked example at the input 3.
Fill in the blanks
\frac6___ = x + 3 \;(x \ne 3) \;\Longrightarrow\; \text___ f(3) = ___
Why: The repair value must equal the limit, which is 6. Any other choice leaves the value and the limit disagreeing, so exactly one repair exists — which is what makes a removable discontinuity removable in a well-defined way.
Matching
Each kind has its own pattern of one-sided limits.
Match the pairs
Why: The first and last rows differ only in the final clause, which is the whole content of the classification: a removable discontinuity has everything continuity needs except the agreement of the value. That is why one value change fixes it and nothing fixes the other two.
Prediction
Commit before reasoning.
Predict first
Can changing the value at a single point repair a jump discontinuity?
Correct: No. The failure is in condition two, and the value has no bearing on the limit.
\[ \text{the limit uses } x \ne a \;\Longrightarrow\; \text{changing } f(a) \text{ cannot change it} \]
Why: Section 2.2 established that a limit is computed entirely from inputs other than the point, so changing the value at the point cannot create a limit where none exists. Choosing the average leaves the value disagreeing with both one-sided limits rather than one — it is worse, not better. Repair is available exactly when the limit already exists and only the value is wrong or missing, which is the definition of removable.
Section
Section 3
Concept
A function is continuous on an open interval if it is continuous at each point of it. On a closed interval, the endpoints are treated with one-sided continuity, since only one side of each lies in the interval at all.
continuity on an interval — Continuous on an open interval means continuous at every point of it. Continuous on a closed interval additionally requires right-continuity at the left endpoint and left-continuity at the right endpoint.
\[ f \text{ continuous on } [a,b]: \; \lim_{x \to a^+}f = f(a), \; \lim_{x \to b^-}f = f(b) \]
The one-sided treatment at endpoints is not a weakening of the definition but the only sensible reading of it. Demanding a two-sided limit at the left endpoint would be asking about inputs the function does not accept.
Figure (svg): Continuity on a closed interval, with one-sided continuity required at the two endpoints
OpenStax Calculus Volume 1, §2.4 Continuity §2.4, pp. 162-164 — continuity over an interval
Picture it
A function continuous on a closed interval.
Figure (svg): Continuity on a closed interval, with one-sided continuity required at the two endpoints
At every interior point both sides are available and both are required. At each endpoint only the inward side exists, and only that side is demanded — which is exactly why the square root is continuous on its whole domain despite the endpoint at zero.
Worked example
Example 2.30. Check the interior, then each endpoint.
\[ \text{Show } f(x) = \sqrt{4 - x^2} \text{ is continuous on } [-2, 2]. \]
Identify the domain
Why: The radicand must be non-negative.
\[ 4 - x ^{2} \ge 0,\text{ so } -2 \le x \le 2 \]
Handle the interior with the composite rule
Why: A root of a polynomial, with the radicand positive inside.
\[ \text{continuous on } (-2, 2) \]
Check the left endpoint from the right only
Why: The radicand approaches 0 from above.
\[ \lim = 0 = f(-2) \]
Check the right endpoint from the left only
Why: Same behaviour.
\[ \lim = 0 = f(2) \]
Conclude
Why: All required conditions hold.
\[ \text{continuous on } [-2, 2] \]
Figure (svg): The solution to Worked example continuity on a closed interval shown as a ladder of expressions, one row per legal move
\[ f \text{ is continuous on } [-2, 2] \]
Verify: ask what a two-sided test at an endpoint would even mean
Why: At x equal to negative 2, inputs below negative 2 make the radicand negative, so f is not defined there and a left-hand limit does not exist. Demanding one would make the function fail on a technicality that has nothing to do with its graph, which is unbroken across the whole interval. The one-sided convention is what makes the definition match the picture, and it is why the semicircle counts as continuous on its closed domain.
OpenStax Calculus Volume 1, §2.4 Continuity §2.4, pp. 163-164
Fill the middle
Choosing the constant so the piecewise function from the worked example has no break.
Fill in the blanks
3k - 3 = 9 \;\Longrightarrow\; k = 4
Why: With k equal to 4 the right-hand limit becomes 9, matching the left-hand limit and the value. Setting the two one-sided limits equal is the standard method for choosing a constant that makes a piecewise function continuous.
Worked example
Checkpoint 2.30. The only doubt is at the seam.
\[ \text{For which } k \text{ is } f(x) = \begin{cases} x^2 & x \le 3 \\ kx - 3 & x > 3 \end{cases} \text{ continuous everywhere?} \]
Note both pieces are continuous on their own
Why: A polynomial on each side.
\[ \text{only } x = 3\text{ is in doubt} \]
Compute the left-hand limit and the value
Why: The first piece owns 3.
\[ 9,\text{ and } f(3) = 9 \]
Compute the right-hand limit
Why: The second piece.
\[ 3 k - 3 \]
Set them equal, which is condition three
Why: All three must agree.
\[ 3 k - 3 = 9 \]
Solve
Why: Adding 3 and dividing by 3.
\[ k = 4 \]
Figure (svg): The solution to Worked example continuity of a piecewise function on an interval shown as a ladder of expressions, one row per legal move
\[ k = 4 \]
Verify: check the pieces really meet
Why: With k equal to 4 the second rule is 4x minus 3, which at 3 gives 9 — matching the first piece exactly. Both one-sided limits and the value are all 9, so all three conditions hold and the graph passes through without a break. Any other k leaves a jump of size 3k minus 12, which is the gap the equation was set to close.
OpenStax Calculus Volume 1, §2.4 Continuity §2.4, pp. 164-164
Trap
\[ f(x) = \sqrt{x} \text{ on } [0, \infty) \]
Test continuity at 0 with a two-sided limit
Why: The student looks for the limit from both sides.
\[ \lim_{x \to 0^-}\sqrt{x} \text{ does not exist} \;\Longrightarrow\; \text{discontinuous} \quad \text{(wrong)} \]
The function has no inputs below 0, so asking about them is asking about points outside its domain.
\[ \lim_{x \to 0^+}\sqrt{x} = 0 = f(0) \;\Longrightarrow\; \text{right-continuous at } 0 \]
At a domain endpoint, require only the side that exists
Why: Continuity is about the function's behaviour on its own domain.
The graph of the square root is plainly unbroken on its whole domain, and any definition declaring it discontinuous at 0 would be describing the definition rather than the function. The one-sided convention at endpoints is what makes continuity mean what the picture shows.
Sorting
Interior points two-sided, endpoints one-sided.
Sort into buckets
Sort each claim.
The reciprocal on the half-open interval from 0 to 1 is the instructive case: 0 is NOT in the interval, so the function's misbehaviour there is irrelevant. Continuity on an interval only ever asks about points of that interval.
Two truths and a lie
All three are about intervals.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. The square root of 4 minus x squared is continuous on the closed interval from negative 2 to 2 and is defined nowhere else at all. Continuity is a statement about the function on the set in question, and it neither requires nor implies anything beyond it.
Prediction
Commit before reasoning.
Predict first
Why is only right-continuity required at the left endpoint of a closed interval?
Correct: Because inputs to the left are outside the domain, so a two-sided limit asks about nothing.
\[ f \text{ right-continuous at } a: \; \lim_{x \to a^+}f(x) = f(a) \]
Why: Continuity describes how a function behaves on its own domain, and there is nothing to the left of the left endpoint to behave. Demanding a two-sided limit would declare the square root discontinuous at 0, and the semicircle discontinuous at both ends, despite both graphs being visibly unbroken. The convention makes the definition track the picture, which is what a good definition should do.
Section
Section 4
Concept
Sums, differences, products and quotients of continuous functions are continuous, with the usual exception where a denominator vanishes. A composite is continuous when the inner function is continuous at the point and the outer one is continuous at the inner's output.
continuity of composites — If g is continuous at a and f is continuous at g of a, then the composite is continuous at a. Equivalently, the limit may be moved inside a continuous outer function.
\[ \lim_{x \to a} f(g(x)) = f\!\left(\lim_{x \to a} g(x)\right) \]
The second form is the one used constantly. It says a limit may be slid inside a continuous function, which is what lets you evaluate the limit of a root, or a sine, or an exponential by evaluating the inside first.
Figure (svg): Continuity passing through a composition, and the limit sliding inside the outer function
OpenStax Calculus Volume 1, §2.4 Continuity §2.4, pp. 164-166 — the algebra of continuous functions
Picture it
Continuity passing through a composition.
Figure (svg): Continuity passing through a composition, and the limit sliding inside the outer function
The inner function carries the input to its own limit, and the outer function, being continuous there, carries that limit to the right output. Nothing is lost at either stage, and that is exactly what continuity guarantees.
Worked example
Example 2.32. Evaluate the inside, then apply the outside.
\[ \text{Evaluate } \lim_{x \to 3} \sqrt{x^2 + 7}. \]
Note the inner function is continuous
Why: A polynomial, continuous everywhere.
\[ x ^{2} + 7\text{ continuous} \]
Note the outer is continuous at the inner's limit
Why: The root is continuous on the non-negative numbers.
Slide the limit inside
Why: The composite rule licenses it.
\[ \sqrt{\lim(x ^{2} + 7)} \]
Evaluate the inside, then the outside
Why: Nine plus 7, then the root.
\[ \sqrt{16} = 4 \]
Figure (svg): The solution to Worked example a limit slid inside a root shown as a ladder of expressions, one row per legal move
\[ \lim_{x \to 3}\sqrt{x^2+7} = 4 \]
Verify: check that the outer function's continuity was genuinely needed
Why: The rule required the root to be continuous at 16, which it is. Had the inner limit been negative, the root would not have been defined there and the manoeuvre would have been illegal — so the second condition is doing real work rather than being a formality. Numerically, at x equal to 3.001 the expression gives about 4.00075, confirming the answer.
OpenStax Calculus Volume 1, §2.4 Continuity §2.4, pp. 165-166
Fill the middle
Evaluating the composite limit from the worked example.
Fill in the blanks
\lim_16\sqrt___ = \sqrt___(x^2+7)} = \sqrt___} = 4
Why: The inner limit is 16, and the root is continuous there, so the slide is legitimate and the answer is 4. Had the inner limit been negative, the root would not have been continuous at it and the move would have been illegal.
Worked example
Checkpoint 2.32. No returning to the definition.
\[ \text{Where is } f(x) = \frac{\sin x}{x^2 - 4} \text{ continuous?} \]
Identify the numerator's continuity
Why: Sine is continuous on all real numbers.
Identify the denominator's continuity
Why: A polynomial.
Apply the quotient rule for continuity
Why: Continuous wherever the denominator is non-zero.
\[ \text{need } x ^{2} - 4 \ne 0 \]
Solve for the exclusions
Why: Factor.
\[ x \ne 2\text{ and } x \ne - 2 \]
State the answer
Why: Everything else.
\[ \text{continuous on all reals except } +- 2 \]
Figure (svg): The solution to Worked example building continuity from known pieces shown as a ladder of expressions, one row per legal move
\[ \text{continuous on } \mathbb{R} \setminus \{-2, 2\} \]
Verify: classify the two breaks
Why: At x equal to 2 the numerator is sine of 2, about 0.909, which is non-zero — so the quotient is unbounded and the break is infinite. The same at negative 2. Neither is removable, since neither numerator vanishes to cancel the factor. Note how little work this took: no limits were computed at all, because the algebra of continuous functions did the whole job from known facts about sine and polynomials.
OpenStax Calculus Volume 1, §2.4 Continuity §2.4, pp. 166-166
Error analysis
A student evaluates a limit through a composition.
Annotate
On: \( \lim_{x \to 0} \frac{1}{g(x)} = \frac{1}{\lim_{x \to 0} g(x)} \quad \text{where } g(x) \to 0 \)
The composite rule has two hypotheses and the second is the one that gets skipped. Always check where the inner limit lands and whether the outer function is continuous there — it is the same check as the quotient law's non-zero denominator, in a different costume.
Sorting
Check where the inner limit lands and whether the outer is continuous there.
Sort into buckets
Sort each attempt.
The sine and exponential cases are safe for any inner limit at all, because those functions are continuous on the whole real line. Roots, reciprocals and logarithms each have a boundary, and the rule's second hypothesis is precisely a check against it.
Matching
Continuity comes free once you know the family.
Match the pairs
Why: Every one of these is continuous on its whole domain, which is the general fact worth carrying: all the standard functions of Chapter 1 are continuous wherever they are defined. Combining them by the algebra of continuity then settles almost every function you will meet without computing a single limit.
Prediction
Commit before reasoning.
Predict first
Why is the composite rule stated as the limit moving inside the outer function?
Correct: Because it converts a hard limit into an easy evaluation.
\[ \lim_{x \to a}f(g(x)) = f\!\left(\lim_{x \to a}g(x)\right) \quad \text{when } f \text{ is continuous there} \]
Why: Instead of analysing the whole composite, you evaluate the inner limit — usually a polynomial, so substitution — and then apply the outer function to that single number. That is the practical content of the rule, and it is used in essentially every limit involving a root, a sine, a logarithm or an exponential from here to the end of the course. Compositions are certainly not always continuous, which is exactly why the rule carries two hypotheses.
Section
Section 5
Concept
If a function is continuous on a closed interval and takes two different values at the ends, then it takes every value in between at some point inside. In particular, a sign change across the interval guarantees a root.
Intermediate Value Theorem — If f is continuous on the closed interval from a to b and z lies between f of a and f of b, then there is at least one c in the open interval with f of c equal to z.
\[ f \text{ continuous on } [a,b], \; f(a) < 0 < f(b) \;\Longrightarrow\; \exists c: f(c) = 0 \]
This is an existence theorem: it guarantees a c exists and says nothing about finding it, nor about how many there are. That combination — certainty about existence, silence about location — is characteristic of the theorems continuity unlocks.
Figure (svg): A continuous curve crossing a horizontal level between two endpoints of opposite sign
OpenStax Calculus Volume 1, §2.4 Continuity §2.4, pp. 166-168 — the Intermediate Value Theorem
Picture it
A continuous curve with a sign change.
Figure (svg): A continuous curve crossing a horizontal level between two endpoints of opposite sign
Somewhere between the two marked points the curve must cross the horizontal axis, because it cannot get from below to above without passing through. The theorem does not say where, and the picture shows why it cannot.
Worked example
Example 2.34. Two evaluations settle existence.
\[ \text{Show that } x^3 - 2x - 1 = 0 \text{ has a root between } 1 \text{ and } 2. \]
Confirm continuity on the closed interval
Why: A polynomial is continuous everywhere.
\[ \text{continuous on } [1, 2] \]
Evaluate at the left endpoint
Why: One minus 2 minus 1.
\[ f(1) = -2 \]
Evaluate at the right endpoint
Why: Eight minus 4 minus 1.
\[ f(2) = 3 \]
Note that zero lies between them
Why: Negative 2 is below 0 and 3 is above.
\[ 0\text{ is intermediate} \]
Apply the theorem
Why: Some interior input gives 0.
Figure (svg): A continuous curve crossing a horizontal level between two endpoints of opposite sign
\[ \exists c \in (1,2): \; c^3 - 2c - 1 = 0 \]
Verify: narrow it by bisecting, and note what the theorem did not give
Why: At the midpoint 1.5 the value is 3.375 minus 3 minus 1, or negative 0.625 — still negative, so the root lies in the interval from 1.5 to 2. Repeating narrows it toward about 1.618. The theorem established that the hunt is worth starting; it gave no root and no method. That division of labour is exactly right, and Section 4.9's Newton's method is the tool for the second half.
OpenStax Calculus Volume 1, §2.4 Continuity §2.4, pp. 167-168
Two truths and a lie
All three are about the theorem.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. The theorem is a one-way implication: a sign change guarantees a root, but its absence guarantees nothing. The square of x minus 1 on the interval from 0 to 2 is positive at both ends and has a root at 1, where it touches the axis without crossing. Reading the implication backwards is the commonest over-extension of this theorem.
Worked example
Checkpoint 2.34. Remove the hypothesis and the conclusion collapses.
\[ \text{Does } f(x) = \frac{1}{x} \text{ take the value } 0 \text{ somewhere on } [-1, 1]? \]
Evaluate at the endpoints
Why: Negative 1 and 1.
\[ f(-1) = -1, f(1) = 1 \]
Note that 0 lies between them
Why: The sign changes across the interval.
\[ 0\text{ is intermediate} \]
Check the theorem's hypothesis
Why: The function is undefined at 0.
\[ NOT\text{ continuous on } [-1, 1] \]
Conclude the theorem does not apply
Why: And in fact no such input exists.
\[ \frac{1}{x}\text{ is never } 0 \]
Figure (svg): A discontinuous function that skips a value, showing the theorem's hypothesis is essential
\[ \tfrac{1}{x} \text{ is never } 0; \text{ the hypothesis fails} \]
Verify: confirm the conclusion genuinely fails, not just the hypothesis
Why: Setting 1 over x equal to 0 gives 1 equal to 0, which is false, so there is genuinely no such input — the function jumps from arbitrarily negative to arbitrarily positive across the asymptote without ever taking the value 0. This is not a case where the theorem merely fails to apply while the conclusion happens to hold; the conclusion is false. That is what makes continuity an essential hypothesis rather than a convenience.
OpenStax Calculus Volume 1, §2.4 Continuity §2.4, pp. 168-168
Trap
\[ f \text{ continuous}, \; f(a) < 0 < f(b) \]
Conclude there is exactly one root in the interval
Why: The student reads the guarantee as a count.
\[ \text{so there is one root in } (a,b) \quad \text{(over-read)} \]
The theorem promises at least one. A continuous function can cross the axis three times, or seventeen, between the same two endpoints.
\[ \exists c \in (a,b): f(c) = 0 \quad \text{- at least one} \]
Read the guarantee as existence, not as a count or a location
Why: The theorem is silent on how many and on where.
The converse over-reading is also common: no sign change does NOT mean no root. A parabola touching the axis at its vertex has a root with no sign change at all, and a curve dipping below and back up between the endpoints has two. Getting a definite answer about the number of roots needs the derivative, which Chapter 4 supplies.
Fill the middle
Applying the theorem to the cubic from the worked example.
Fill in the blanks
f(1) = -2 < 0 < 3 = f(2) \;\Longrightarrow\; \exists c \in (1,2) \text0 f(c) = ___
Why: Zero lies between the two endpoint values, so the theorem places a root strictly inside the interval. The two evaluations are the entire proof; nothing further is needed to establish existence.
Ranking
Using the theorem to show a root exists.
Put in order
Why: Step a is the one that is skipped, and the reciprocal example shows the cost: with a discontinuity in the interval the conclusion can be outright false. Step e is separated deliberately, because the theorem's job ends at step d — it establishes existence and hands the search to a different tool.
Prediction
Commit before reasoning.
Predict first
You show a continuous function changes sign on [1, 2]. What do you now know?
Correct: At least one root lies strictly inside, and nothing more.
\[ \text{existence: yes} \quad \text{location: no} \quad \text{count: no} \]
Why: The theorem is purely an existence statement. It gives no location, no count, and no information about whether the function rises or falls — a wildly oscillating continuous function satisfies it just as well as a straight line. What makes it valuable is that two evaluations settle whether a search is worth starting at all, which is a genuinely useful thing to know before committing to a numerical method that might otherwise hunt for something that is not there.
Comparison
Fill the blanks. The one-sided limits decide everything.
Comparison matrix
| Kind | One-sided limits | Repairable? |
|---|---|---|
| Removable | both exist and agree | yes: set the value equal to the limit |
| Jump | both exist and differ | no: no single value satisfies both sides |
| Infinite | at least one is unbounded | no: there is no finite value to approach |
| None: continuous | both agree and match the value | nothing to repair |
Only the top row can be fixed, and its repair value is forced to be the limit. That is the practical content of the classification, and it is decided entirely without looking at the value at the point.
Pattern
Given a function and a point, decide continuity and classify any failure.
Step five is the one that saves the most time. Almost no function in this course needs a point-by-point check; naming the families it is built from and where the denominators vanish settles it in one line.
Stewart, Calculus: Early Transcendentals 8e, §2.5 Continuity §2.5, pp. 114-125
Check
All three conditions.
Check your understanding
For f equal to (x^2-4)/(x-2) except f(2) = 3, is f continuous at 2?
Answer: A
Why: The value and the limit both exist but differ, so the third condition fails.
Check
Classification. Compute both one-sided limits.
Check your understanding
Classify the discontinuity of |x-1|/(x-1) at x = 1.
Answer: A
Why: The one-sided limits are -1 and 1: both exist and differ, which is a jump.
Check
The theorem. Check the hypothesis first.
Check your understanding
f is continuous on [1,2] with f(1) = -2 and f(2) = 3. What follows?
Answer: A
Why: Zero lies between -2 and 3, so the theorem places at least one root inside the interval.
Real world
A thermostat controls a room. At 6 a.m. the temperature is 16 degrees; by 10 a.m. it is 22 degrees. Separately, a company's share price closed at 40 dollars on Monday and 55 dollars on Tuesday.
Discussion prompt
For which of these can you guarantee the quantity passed through exactly 19 degrees, or exactly 47 dollars, at some moment? Justify using the theorem, and say precisely what the difference is.
Hint: Ask whether each quantity varies continuously in time.
Answer:
The temperature: yes. Room temperature is a continuous function of time — it cannot get from 16 to 22 without passing through every value between, because heat transfer is a continuous physical process. The theorem applies and guarantees a moment at exactly 19 degrees.
\[ T \text{ continuous on } [6, 10], \; T(6) = 16 < 19 < 22 = T(10) \;\Longrightarrow\; \exists t: T(t) = 19 \]
The share price: no. Prices move in discrete ticks and the market closes overnight; the price can gap from 44 to 49 between sessions without ever trading at 47. The function is not continuous, so the hypothesis fails — and unlike the temperature, the conclusion genuinely can be false.
This is the same distinction as the reciprocal example: the theorem is not merely inapplicable to the discontinuous case, it is false there. An overnight gap is a jump discontinuity, and a jump is precisely a step over intermediate values.
Note what the theorem still does not give for the temperature: it does not say WHEN the room was at 19 degrees, nor whether that happened once or several times if the heating cycled. Existence is guaranteed; everything else needs more information.
Commit first
Answer, then rate your confidence honestly.
Predict first
Which discontinuity can be repaired by changing the function at a single point?
Correct: A removable one — set the value equal to the limit.
\[ \text{limit exists} \iff \text{repairable, and the repair value is the limit} \]
Why: Repair is possible exactly when the limit already exists, because the limit is computed from inputs other than the point and no change at the point can affect it. For a jump, the two sides disagree with each other, so any single value disagrees with at least one; averaging makes it disagree with both. For an infinite discontinuity there is no finite value being approached at all. The classification is precisely a statement about which repairs are available.
Explain it
They think the Intermediate Value Theorem is obvious and therefore useless.
Discussion prompt
In four sentences or fewer, explain what the theorem gives you that is not obvious, using an example where it does real work.
Hint: Ask them to find a root of the cubic by algebra first.
Answer:
Ask them to solve x cubed minus 2x minus 1 equals 0 by algebra. There is no factoring that works and no formula they know, and after ten minutes they still cannot say whether a root between 1 and 2 exists at all.
Two substitutions settle it: the function is negative at 1 and positive at 2, so a root is there. That is the theorem doing real work — it converts an unsolvable equation into two arithmetic evaluations, and it tells you a numerical hunt will succeed before you start one. What is obvious for a picture you have already drawn is not obvious for a function you cannot draw.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For the three conditions, write them as a numbered checklist and work down it. For classification, always compute both one-sided limits first — the value at the point never decides the kind. For endpoints, require only the side that is inside the domain. For the theorem, say 'at least one, somewhere' out loud every time, and check continuity before anything else. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
At the top, write the three conditions for continuity as a numbered list, and beside each draw a small graph of a function failing that condition and only that one. Below, make a three-column table of the discontinuity types: for each, write the pattern of one-sided limits, an example function with its point, and whether it can be repaired. In the middle of the page, take the quotient of x squared minus 9 by x minus 3, show the cancellation, and state the value that repairs it at 3. Then take the piecewise function that is x squared up to 3 and kx minus 3 above, and find the k that makes it continuous, showing the equation you set. At the bottom, draw a continuous curve that is negative at 1 and positive at 2, mark a root, and write the theorem's statement beside it; then draw a second curve with a jump that is negative at 1 and positive at 2 and never zero, and write one sentence saying which hypothesis it violates. In a margin, write the composite rule in the form where the limit slides inside, and note its two hypotheses.
If your three failure graphs at the top all look like the same picture, look again — one should have a hole, one a step, and one a dot floating off the curve, and each should satisfy the other two conditions perfectly.
Recap
Five things, and the last is the first theorem in this course that promises something exists without producing it.
| If you see | Then |
|---|---|
| A hole in the graph | Removable: define the value as the limit |
| A step in the graph | Jump: not repairable |
| A vertical asymptote | Infinite: not repairable |
| An endpoint of the domain | Require one-sided continuity only |
| A combination of standard functions | Continuous except where a denominator vanishes |
| A continuous sign change | At least one root, location unknown |
| No sign change | Nothing follows either way |
Section 2.5 closes the chapter by making the word 'approaches' precise. The epsilon-delta definition is what all of Chapter 2 has been using informally, and it is what turns every result here from plausible into proved.
OpenStax Calculus Volume 1, §2.4 Continuity §2.4, pp. 155-168 — everything on these slides traces back here
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