This deck gives the three-part definition of continuity at a point and uses it to classify removable, jump, and infinite discontinuities. It then covers continuity on intervals and for the standard function families, composites and passing a limit inside, and choosing parameters that make a piecewise function continuous, before closing with the Intermediate Value Theorem and bisection. It targets the students who check the limit but forget that the function must be defined, who think cancelling a factor erases the hole, and who use the IVT without checking continuity or read it as guaranteeing exactly one root.
Subject: Calculus I · 132 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Title
Calculus I - Deck 03
No holes, no jumps, no asymptotes - and what an unbroken curve is forced to do.
Objectives
Continuity is the bridge between limits and everything that follows. By the end of this deck you can:
Everything here rests on the limit skills from the previous two decks. If a limit computation ever feels shaky, that is the thing to shore up first.
Warm-up
Discussion prompt
Before we open Continuity and the Intermediate Value Theorem: without looking back, what was the main idea of Limit Laws and Computing Limits Algebraically, and what could you do by the end of it that you could not do before?
Hint: One sentence for the idea, one for the skill. If the second one is blank, that is the part to revisit.
Answer:
That deck covers the limit laws and the conditions attached to each of them, then works out when direct substitution is legal and what to do when it is not. It handles the zero-over-zero indeterminate form with factoring and cancelling, conjugates, and complex fractions, takes piecewise limits at the seam, and finishes with the Squeeze Theorem and the classic sine-over-x limit. It targets the traps of cancelling a factor and forgetting the hole, treating zero over zero as automatically zero or one, using the quotient law when the bottom limit is zero, and multiplying by a conjugate on only part of the fraction.
Section
Part 1
Intuition
Here is the picture everyone starts with: a function is continuous on a stretch if you can draw its graph there without lifting your pencil.
That single image already contains every failure mode. To lift the pencil you must do one of three things: skip a single point, hop to a new height, or run off the page.
Those three are exactly the hole, the jump, and the asymptote. The rest of this deck just makes that precise enough to compute with.
The pencil picture is a great intuition and a bad definition - some genuinely continuous functions wiggle too fast to draw. So we replace it with a limit statement.
Counterexample
Discussion prompt
The pencil picture is a great intuition and a bad definition - some genuinely continuous functions wiggle too fast to draw. So we replace it with a limit statement.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Picture it
Animation
Shows: The pencil never leaves the paper — a rendered Manim animation.
Rendered with Manim.
Takeaway: An informal picture of continuity — useful, and not a definition.
Concept
We do not ask "is this function continuous?" as a yes-or-no question about the whole function. We ask it one input at a time.
\[ f \text{ is continuous at } x = a \]
Continuity on an interval is then just: continuous at every point of that interval. Local first, global second.
Analogy
Discussion prompt
Explain Continuity is a statement about one point by analogy to something with no Calculus I in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
We do not ask "is this function continuous?" as a yes-or-no question about the whole function. We ask it one input at a time.
Concept
A function is continuous at a point when three separate things are all true. Miss any one and continuity fails.
\[ \textbf{(1)}\ \ f(a) \text{ exists} \]
\[ \textbf{(2)}\ \ \lim_{x \to a} f(x) \text{ exists} \]
\[ \textbf{(3)}\ \ \lim_{x \to a} f(x) = f(a) \]
Condition three is the punchline, but it is only meaningful once the first two hold - you cannot compare two things until both of them exist.
continuous at a point — The function is defined there, the two-sided limit there exists, and those two numbers are the same. All three, every time.
Explain it
Discussion prompt
Explain The three-part definition to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
A function is continuous at a point when three separate things are all true. Miss any one and continuity fails.
Picture it
Animation
Shows: The three conditions defining continuity at a point.
Three separate things that must all hold.
Takeaway: The value must exist, the limit must exist, and the two must match. Each can fail independently, and each failure looks different on a graph.
Concept
The first condition is the one students skip. Before you compute any limit, ask: does the function actually have a value at this input?
\[ f(a) \text{ must be a real number} \]
A denominator of zero, a square root of a negative, a logarithm of zero or less, or a piecewise rule that simply does not cover the point - any of these kills condition one on the spot.
\[ f(x) = \frac{x^2 - 4}{x - 2} \quad \Rightarrow \quad f(2) = \frac{0}{0} \ \text{ undefined} \]
Concept
The second condition is the limit work from the last two decks, unchanged. The limit exists exactly when both one-sided limits exist and agree.
\[ \lim_{x \to a^{-}} f(x) = \lim_{x \to a^{+}} f(x) = L \]
Notice what this condition does not care about: the value at the point itself. The limit is a statement about the neighbors.
Picture it
Animation
Shows: Condition two: the limit must exist — a rendered Manim animation.
Rendered with Manim.
Takeaway: A jump fails here, however nicely the function is defined at the point.
Picture it
Animation
Shows: Condition one: the value must exist — a rendered Manim animation.
Rendered with Manim.
Takeaway: A hole fails here even when the limit is perfectly well behaved.
Concept
Now you have two independently computed numbers: the height the neighbors are heading toward, and the height actually posted at the point.
\[ \lim_{x \to a} f(x) = f(a) \]
Continuity says the graph keeps its promise. The neighbors predict a height, and the point delivers exactly that height.
When a function is continuous at a point, evaluating the limit is the same as plugging in. That is the whole reason direct substitution ever worked.
Picture it
Animation
Shows: Condition three: the two must match — a rendered Manim animation.
Rendered with Manim.
Takeaway: Both can exist and still disagree — a removable break.
Intuition
The three conditions are not arbitrary. Each one blocks a different way the pencil could lift.
| Condition that fails | What the graph does | Name |
|---|---|---|
| 1 only: no value posted | the curve arrives but the point is missing | removable (hole) |
| 2: sides disagree | the curve hops to a new height | jump |
| 2: sides blow up | the curve runs off the page | infinite |
| 3 only: value posted wrong | the point sits off the curve | removable (misplaced point) |
So the classification you will learn in Part 2 is just bookkeeping on which condition broke.
Comparison
Comparison matrix
From Each condition rules out one kind of break: refill the What the graph does column from what you know. The rest of the table is as it appeared.
| Condition that fails | What the graph does | Name |
|---|---|---|
| 1 only: no value posted | the curve arrives but the point is missing | removable (hole) |
| 2: sides disagree | the curve hops to a new height | jump |
| 2: sides blow up | the curve runs off the page | infinite |
| 3 only: value posted wrong | the point sits off the curve | removable (misplaced point) |
Picture it
Animation
Shows: Removable, jump and infinite discontinuities listed.
Only the first one is repairable.
Takeaway: Only the removable break can be repaired by redefining a single point. A jump and an asymptote are damage no single value can fix.
Ranking
Put in order
Put the moves of Worked example: continuity where there is a hole into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Substituting two gives zero over zero, which is not a number.
Worked example
Decide whether this function is continuous at the input two, and if not, say exactly which condition fails.
\[ f(x) = \frac{x^2 - 4}{x - 2}, \qquad a = 2 \]
Test condition 1 first: evaluate at the point
Why: Substituting two gives zero over zero, which is not a number. The function has no value at two, so condition one already fails.
Test condition 2 anyway: compute the limit
Why: Failing condition one does not stop the limit from existing - and the answer tells us what kind of break this is. Factor the numerator.
\[ \frac{x^2-4}{x-2} = \frac{(x-2)(x+2)}{x-2} = x + 2 \quad (x \neq 2) \]
Take the limit of the simplified expression
Why: For every input except two the two expressions agree, and the limit never looks at the point itself.
\[ \lim_{x \to 2} f(x) = \lim_{x \to 2} (x+2) = 4 \]
State the verdict
Why: The limit exists and equals four, but there is nothing at the point to compare it to. Condition one fails, so the function is not continuous at two.
\[ f \text{ is discontinuous at } x = 2 \ \text{(removable)} \]
Verify with a table of nearby values
Why: At input 1.9 the quotient is negative 0.39 over negative 0.1, which is 3.9. At input 2.1 it is 0.41 over 0.1, which is 4.1. The neighbors close in on four while the point itself stays empty - exactly a hole.
| input | value of f | distance from 4 |
|---|---|---|
| 1.9 | 3.9 | 0.1 |
| 1.99 | 3.99 | 0.01 |
| 2 | undefined | - |
| 2.01 | 4.01 | 0.01 |
| 2.1 | 4.1 | 0.1 |
Picture it
Animation
Shows: Each line of the worked example "continuity where there is a hole", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At input 1.9 the quotient is negative 0.39 over negative 0.1, which is 3.9. At input 2.1 it is 0.41 over 0.1, which is 4.1. The neighbors close in on four while the point itself stays empty - exactly a hole.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The student computes the limit, gets a clean number, and declares continuity.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: This part is correct - the limit really is four.
Run the checklist in order, starting with the value.
Why: This part is correct - the limit really is four.
Trap
The student computes the limit, gets a clean number, and declares continuity.
\[ f(x) = \frac{x^2-4}{x-2} \]
Cancel and take the limit: the answer is four
Why: This part is correct - the limit really is four.
Conclude the function is continuous at two
Why: Wrong. Only condition two was checked. Nobody asked whether the function has a value at two.
\[ f(2) = \frac{0}{0} \ \text{ does not exist} \]
Run the checklist in order, starting with the value.
\[ f(x) = \frac{x^2-4}{x-2} \]
Condition 1: is the function defined at two?
Why: No - the denominator is zero there. Continuity is already dead; keep going only to name the break.
Condition 2: the limit is four, so the break is a hole
Why: A limit that exists while the value does not is precisely a removable discontinuity. Correct verdict: discontinuous at two, removable.
\[ \lim_{x \to 2} f(x) = 4 \ne f(2) \ (\text{undefined}) \]
Notation
Annotate
From Trap: "the limit exists, so it is continuous" — read this one piece at a time. What is each part doing?
On: \( f(x) = \frac{x^2-4}{x-2} \)
Step zero
Discussion prompt
Worked example: a piecewise function at its seam — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Condition 1: evaluate at one
Answer:
Worked example
Test this function for continuity at the input one - the only place its rule changes.
\[ f(x) = \begin{cases} x^2 + 1, & x < 1 \\ 3x - 1, & x \ge 1 \end{cases} \]
Condition 1: evaluate at one
Why: The input one satisfies the second case, so use the second rule: three times one minus one is two.
\[ f(1) = 3(1) - 1 = 2 \]
Condition 2, left side: use the first rule
Why: Approaching from below means every input is less than one, so the first rule applies the whole way in.
\[ \lim_{x \to 1^{-}} f(x) = 1^2 + 1 = 2 \]
Condition 2, right side: use the second rule
Why: Approaching from above means every input is at least one, so the second rule applies. Both one-sided limits are two, so the two-sided limit exists and equals two.
\[ \lim_{x \to 1^{+}} f(x) = 3(1) - 1 = 2 \]
Condition 3: compare
Why: The limit is two and the value is two. All three conditions hold, so the function is continuous at one - and everywhere else too, since each piece is a polynomial.
\[ \lim_{x \to 1} f(x) = 2 = f(1) \]
Verify numerically just off the seam
Why: At input 0.99 the first rule gives 0.9801 plus 1, which is 1.9801. At input 1.01 the second rule gives 3.03 minus 1, which is 2.03. Both sit within a few hundredths of two, and they close in from opposite sides - no jump.
| input | rule used | value |
|---|---|---|
| 0.99 | first | 1.9801 |
| 0.999 | first | 1.998001 |
| 1 | second | 2 |
| 1.001 | second | 2.003 |
| 1.01 | second | 2.03 |
Section
Part 2
Concept
A break is removable when the two-sided limit exists but the point does not match it - either because nothing is there, or because a stray value was posted somewhere else.
\[ \lim_{x \to a} f(x) = L \ \text{ exists}, \quad \text{but } f(a) \ne L \ \text{ or } f(a) \ \text{undefined} \]
Figure (svg): A straight rising line with a small open circle where one point has been removed
It is called removable because you could patch it by redefining the function at that one input to be the limit. One repair, one point.
Picture it
Figure (svg): Two horizontal segments at different heights meeting at one input, with an open circle on the lower branch and a filled dot on the upper
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
A break is a jump when both one-sided limits exist as finite numbers but they are different. The curve arrives at two different heights depending on which way you come.
Concept
A break is a jump when both one-sided limits exist as finite numbers but they are different. The curve arrives at two different heights depending on which way you come.
\[ \lim_{x \to a^{-}} f(x) = L_1, \quad \lim_{x \to a^{+}} f(x) = L_2, \quad L_1 \ne L_2 \]
Figure (svg): Two horizontal segments at different heights meeting at one input, with an open circle on the lower branch and a filled dot on the upper
A jump is not removable. Changing one value cannot reconcile two different one-sided limits - you would have to bend the whole graph.
Picture it
Animation
Shows: A jump, seen — a rendered Manim animation.
Rendered with Manim.
Takeaway: Two different heights on either side, with nothing in between.
Picture it
Figure (svg): A dashed vertical line with one branch of a curve rising steeply toward it from the left and another branch falling from below on the right
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
A break is infinite when at least one of the one-sided limits grows without bound. The graph has a vertical asymptote there.
Concept
A break is infinite when at least one of the one-sided limits grows without bound. The graph has a vertical asymptote there.
\[ \lim_{x \to a^{-}} f(x) = \pm\infty \quad \text{or} \quad \lim_{x \to a^{+}} f(x) = \pm\infty \]
Figure (svg): A dashed vertical line with one branch of a curve rising steeply toward it from the left and another branch falling from below on the right
Saying a limit equals infinity is shorthand for how the limit fails, not a claim that it exists. So condition two fails, and the discontinuity is not removable.
Intuition
You never have to guess the type. Compute the limit from the left and the limit from the right, then read the answer off this table.
| Left limit | Right limit | Type of break |
|---|---|---|
| a number L | the same number L | removable (or no break at all) |
| a number | a different number | jump |
| unbounded | anything | infinite |
| anything | unbounded | infinite |
| fails to settle | fails to settle | essential (oscillating) |
The last row is the wild case from deck one - the sine of one over the input near zero. It is not removable, not a jump, and not an asymptote; it simply never settles.
Trade off
Comparison matrix
From Which break is it? Ask the two one-sided limits: every row here is a choice with a cost. Fill the Right limit column, then say which row you would actually pick and what you give up for it.
| Left limit | Right limit | Type of break |
|---|---|---|
| a number L | the same number L | removable (or no break at all) |
| a number | a different number | jump |
| unbounded | anything | infinite |
| anything | unbounded | infinite |
| fails to settle | fails to settle | essential (oscillating) |
Estimation
Predict first
Find every discontinuity and name its type.
Commit before you compute: what does Worked example: classify every break of a rational function come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify all three claims numerically
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. At input 2.01 the quotient is 0.0101 over 0.0401, about 0.2519 - closing in on one quarter.
Worked example
Find every discontinuity and name its type.
\[ f(x) = \frac{x^2 - 3x + 2}{x^2 - 4} \]
Factor top and bottom completely
Why: For a rational function every break sits where the denominator is zero, and factoring is what tells you whether the numerator vanishes there too.
\[ f(x) = \frac{(x-1)(x-2)}{(x-2)(x+2)} \]
List the candidate inputs: two and negative two
Why: Those are the zeros of the denominator; the function is defined and continuous everywhere else because it is a quotient of polynomials.
At the input two, the factor cancels
Why: The shared factor cancels for every input except two itself, so the limit comes from the reduced expression. The limit exists, but the original function is undefined there - a removable break.
\[ \lim_{x \to 2} f(x) = \lim_{x \to 2} \frac{x-1}{x+2} = \frac{1}{4} \]
At the input negative two, nothing cancels
Why: The numerator there is four plus six plus two, which is twelve, while the denominator goes to zero. A nonzero number over a vanishing denominator blows up - an infinite break.
\[ \text{numerator at } -2:\ (-3)(-4) = 12 \ne 0 \]
Pin down the sign on each side of negative two
Why: Just left of negative two both factors of the denominator are negative, so their product is positive and the quotient is large and positive. Just right of it, the factor of x plus two turns positive while x minus two stays negative, flipping the sign.
\[ \lim_{x \to -2^{-}} f(x) = +\infty, \qquad \lim_{x \to -2^{+}} f(x) = -\infty \]
Verify all three claims numerically
Why: At input 2.01 the quotient is 0.0101 over 0.0401, about 0.2519 - closing in on one quarter. At input negative 2.01 it is 12.0701 over 0.0401, about positive 301. At negative 1.99 it is 11.9301 over negative 0.0399, exactly negative 299. Removable at two, infinite at negative two, confirmed.
| input | value of f | reading |
|---|---|---|
| 1.99 | 0.24813 | heading to one quarter |
| 2.01 | 0.25187 | heading to one quarter |
| -2.01 | 301.0 | blowing up positive |
| -1.99 | -299.0 | blowing up negative |
Picture it
Animation
Shows: Each line of the worked example "classify every break of a rational function", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At input 2.01 the quotient is 0.0101 over 0.0401, about 0.2519 - closing in on one quarter. At input negative 2.01 it is 12.0701 over 0.0401, about positive 301. At negative 1.99 it is 11.9301 over negative 0.0399, exactly negative 299. Removable at two, infinite at negative two, confirmed.
Pattern
A four-move routine that settles any break you will meet in this course.
If the value exists, the limit exists, and they are equal, there was no break at all - the suspect was innocent.
| What you found | Verdict |
|---|---|
| value missing, limit exists | removable |
| value present but different from the limit | removable |
| one-sided limits differ, both finite | jump |
| a one-sided limit is unbounded | infinite |
| value equals the limit | continuous |
Elimination
Eliminate the wrong options
What kind of discontinuity does this function have at the input five?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: The denominator factors as x minus five times x plus five, so the shared factor cancels and the function equals one over x plus five for every input except five. The limit is therefore one tenth, but the original function is undefined at five, so the break is removable.
Check
Factor before you decide. Work it on paper first.
\[ f(x) = \frac{x-5}{x^2-25}, \qquad \text{at } x = 5 \]
Check your understanding
What kind of discontinuity does this function have at the input five?
Answer: A
Why: The denominator factors as x minus five times x plus five, so the shared factor cancels and the function equals one over x plus five for every input except five. The limit is therefore one tenth, but the original function is undefined at five, so the break is removable.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The student simplifies and then treats the simplified formula as the whole story.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The algebra is right, but an equals sign has been claimed for every input, including three.
Carry the restriction along with the cancellation.
Why: The algebra is right, but an equals sign has been claimed for every input, including three.
Trap
The student simplifies and then treats the simplified formula as the whole story.
\[ f(x) = \frac{x^2-9}{x-3} \]
Cancel the shared factor and rewrite the function as a line
Why: The algebra is right, but an equals sign has been claimed for every input, including three.
\[ f(x) = x + 3 \quad \text{(claimed for all } x) \]
Conclude the graph is a line and is continuous everywhere
Why: Wrong. Cancelling divided by x minus three, which is only legal when that quantity is not zero. The graph of the original still has a missing point.
\[ f(3) = \frac{0}{0} \ \text{ still undefined} \]
Carry the restriction along with the cancellation.
\[ f(x) = \frac{x^2-9}{x-3} \]
Cancel, and write down where the identity is valid
Why: The two expressions agree at every input except three; at three one is undefined and the other is six. That restriction is part of the answer.
\[ f(x) = x + 3, \quad x \ne 3 \]
Report a line with a hole at the point three, six
Why: The limit at three is six and the value is missing, so the break is removable. Cancelling finds the limit; it does not repair the function.
\[ \lim_{x \to 3} f(x) = 6, \qquad f(3) \ \text{undefined} \]
Notation
Annotate
From Trap: "it cancelled, so the hole is gone" — read this one piece at a time. What is each part doing?
On: \( \lim_{x \to 3} f(x) = 6, \qquad f(3) \ \text{undefined} \)
Section
Part 3
Concept
Once you can test one point, an interval is easy: a function is continuous on an open interval when it is continuous at every point inside it.
\[ f \ \text{continuous on } (a,b) \iff f \ \text{continuous at every } c \in (a,b) \]
Open intervals are the comfortable case: every point has neighbors on both sides, so the ordinary two-sided limit makes sense everywhere.
Concept
At the left end of a domain there is nothing to the left. Demanding a two-sided limit there would be unfair, so we only ask for the side that exists.
\[ \text{right-continuous at } a: \quad \lim_{x \to a^{+}} f(x) = f(a) \]
\[ \text{left-continuous at } b: \quad \lim_{x \to b^{-}} f(x) = f(b) \]
one-sided continuity — The same three conditions, but with a one-sided limit in place of the two-sided one. It is what continuity means at the edge of a domain.
Concept
Putting those together gives the phrase you will see in the hypotheses of every big theorem for the rest of the course.
\[ f \ \text{continuous on } [a,b] \]
Remember this definition. The Intermediate Value Theorem in Part 6 asks for exactly this, on exactly a closed interval, and it is the hypothesis students skip.
Sorting
Sort into buckets
These are the pieces of Continuity and the Intermediate Value Theorem, out of order. Put each one back under the part of the lesson it belongs to.
Pattern
Predict first
The table runs: 3.01 | 0.01 | 0.1 · 3.0001 | 0.0001 | 0.01 · 3 | 0 | 0
In Worked example: where is a square root continuous?, given the rows so far: what is the next one — the row where input is 2.99?
Correct: 2.99 | -0.01 | undefined
| input | radicand | value |
|---|---|---|
| 3.01 | 0.01 | 0.1 |
| 3.0001 | 0.0001 | 0.01 |
| 3 | 0 | 0 |
| 2.99 | -0.01 | undefined |
Why: The relationship between the columns, not the individual numbers, is what generates the next row. A real square root needs a nonnegative radicand, so the inside must be at least zero.
Worked example
State every interval on which this function is continuous.
\[ g(x) = \sqrt{x-3} \]
Find the domain first
Why: A real square root needs a nonnegative radicand, so the inside must be at least zero. Nothing outside the domain can be continuous, because condition one already fails there.
\[ x - 3 \ge 0 \ \Longrightarrow \ x \ge 3 \]
Handle the interior of the domain
Why: For every input strictly greater than three the square root of a positive number is a composition of continuous pieces, so substitution gives the limit.
\[ \lim_{x \to c} \sqrt{x-3} = \sqrt{c-3} = g(c), \quad c > 3 \]
Handle the endpoint with a one-sided limit
Why: There is no left side to check at three, so the correct question is whether the function is right-continuous there. The right-hand limit is zero and the value is zero.
\[ \lim_{x \to 3^{+}} \sqrt{x-3} = 0 = g(3) \]
Report the interval with the endpoint included
Why: Interior continuity plus right-continuity at the endpoint is exactly continuity on the closed-at-three interval. Writing an open bracket here would throw away a point where the function is perfectly well behaved.
\[ g \ \text{is continuous on } [3,\infty) \]
Verify at the endpoint with nearby values
Why: At input 3.01 the value is the square root of 0.01, which is 0.1. At 3.0001 it is the square root of 0.0001, which is 0.01. These shrink toward zero, matching the value at three exactly.
| input | radicand | value |
|---|---|---|
| 3.01 | 0.01 | 0.1 |
| 3.0001 | 0.0001 | 0.01 |
| 3 | 0 | 0 |
| 2.99 | -0.01 | undefined |
Picture it
Animation
Shows: Each line of the worked example "where is a square root continuous?", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At input 3.01 the value is the square root of 0.01, which is 0.1. At 3.0001 it is the square root of 0.0001, which is 0.01. These shrink toward zero, matching the value at three exactly.
Concept
You do not re-derive continuity from scratch for common functions. These standard families are continuous at every point of their domains, full stop.
| Family | Continuous on | Watch out for |
|---|---|---|
| Polynomials | all real numbers | nothing - they never break |
| Rational functions | everywhere the denominator is nonzero | zeros of the denominator |
| Roots | where the radicand is allowed | even roots of negatives |
| Sine and cosine | all real numbers | nothing |
| Tangent, secant, and friends | where the denominator trig value is nonzero | odd multiples of a quarter turn |
| Exponentials | all real numbers | nothing |
| Logarithms | positive inputs only | zero and negative inputs |
That is the payoff of the limit laws from deck two: for these families, taking a limit is the same as plugging in.
Comparison
Comparison matrix
From The families that are continuous on their domains: refill the Watch out for column from what you know. The rest of the table is as it appeared.
| Family | Continuous on | Watch out for |
|---|---|---|
| Polynomials | all real numbers | nothing - they never break |
| Rational functions | everywhere the denominator is nonzero | zeros of the denominator |
| Roots | where the radicand is allowed | even roots of negatives |
| Sine and cosine | all real numbers | nothing |
| Tangent, secant, and friends | where the denominator trig value is nonzero | odd multiples of a quarter turn |
| Exponentials | all real numbers | nothing |
| Logarithms | positive inputs only | zero and negative inputs |
Concept
The limit laws hand you a closure rule for free. If two functions are continuous at a point, so is almost anything you build from them.
\[ f+g, \quad f-g, \quad cf, \quad fg, \quad \frac{f}{g}\ \ (g(a) \ne 0) \]
The one condition worth memorizing is the quotient's: the bottom function must not be zero at the point. Every other combination is unconditional.
So any formula assembled from the standard families by adding, subtracting, multiplying, and dividing is continuous everywhere it is defined.
Intuition
Here is the practical shortcut that turns most textbook continuity questions into a domain question.
If a function is written as one formula built from the standard families, then it is continuous exactly on its domain. So the work is: find the domain, and you are done.
That shortcut has one honest exception: piecewise definitions. A piecewise rule can have a perfectly fine domain and still jump at a seam, so seams always get checked by hand.
Step zero
Discussion prompt
Worked example: find every point of discontinuity — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Factor the denominator
Answer:
Worked example
Where is this function continuous, and what happens where it is not?
\[ h(x) = \frac{x+1}{x^2 - x - 6} \]
Factor the denominator
Why: This is a rational function, so it is continuous everywhere the bottom is nonzero. Factoring finds those inputs and shows whether anything cancels.
\[ x^2 - x - 6 = (x-3)(x+2) \]
Set each factor to zero
Why: The denominator vanishes only at three and at negative two, so those are the only possible breaks.
\[ x = 3 \quad \text{or} \quad x = -2 \]
Check whether the numerator kills either one
Why: At three the numerator is four, and at negative two it is negative one. Neither is zero, so nothing cancels and both breaks are infinite rather than removable.
\[ \text{numerator: } 3+1 = 4, \qquad -2+1 = -1 \]
State the continuity set
Why: Everywhere else the function is a quotient of polynomials with a nonzero denominator, hence continuous. Write it as the real line with the two bad inputs removed.
\[ (-\infty,-2) \cup (-2,3) \cup (3,\infty) \]
Verify the two bad inputs and one good one
Why: At three the denominator is nine minus three minus six, which is zero. At negative two it is four plus two minus six, which is zero. At zero the denominator is negative six, so the value is negative one sixth - defined, as claimed.
| input | denominator | status |
|---|---|---|
| 3 | 0 | infinite break |
| -2 | 0 | infinite break |
| 0 | -6 | continuous, value is -1/6 |
| 5 | 14 | continuous, value is 3/7 |
Picture it
Animation
Shows: Each line of the worked example "find every point of discontinuity", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At three the numerator is four, and at negative two it is negative one. Neither is zero, so nothing cancels and both breaks are infinite rather than removable.
Hypothesis
Predict first
Worked example: where is the tangent function continuous? is about to be worked. State your hypothesis first: which rule or definition decides this one, and what is the first move it forces? Then watch whether the example agrees with you.
Correct: Write it as a quotient of two continuous functions
Why: Sine and cosine are continuous on the whole real line, so the quotient rule for continuity applies wherever the bottom is nonzero.
A hypothesis you wrote down is falsifiable; a vague sense of how it will go is not. If the example opens somewhere else, that gap is the thing worth chasing.
Worked example
Tangent is the standard example of a function that is continuous on its domain but whose domain has infinitely many gaps.
\[ T(x) = \tan x = \frac{\sin x}{\cos x} \]
Write it as a quotient of two continuous functions
Why: Sine and cosine are continuous on the whole real line, so the quotient rule for continuity applies wherever the bottom is nonzero.
Find where the denominator vanishes
Why: Cosine is zero at a quarter turn and then every half turn after that, in both directions. Those inputs are the only candidates for a break.
\[ \cos x = 0 \iff x = \frac{\pi}{2} + k\pi, \quad k \in \mathbb{Z} \]
Classify those breaks
Why: At each of them sine equals plus or minus one, so the numerator is nonzero while the denominator vanishes. Every break is infinite - a vertical asymptote, never a hole.
\[ \lim_{x \to (\pi/2)^{-}} \tan x = +\infty, \qquad \lim_{x \to (\pi/2)^{+}} \tan x = -\infty \]
Verify with values squeezed up to a quarter turn
Why: A quarter turn is about 1.5708 in radians. At 1.5 the tangent is about 14.10; at 1.57 it is about 1255.8, shooting up. Just past it, at 1.6, the tangent is about negative 34.23 - the sign flip you expect across a vertical asymptote.
| input (radians) | tangent | reading |
|---|---|---|
| 1.5 | 14.10 | climbing |
| 1.57 | 1255.8 | climbing fast |
| 1.6 | -34.23 | on the other branch |
| 1.8 | -4.286 | coming back up |
Picture it
Animation
Shows: Each line of the worked example "where is the tangent function continuous?", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: A quarter turn is about 1.5708 in radians. At 1.5 the tangent is about 14.10; at 1.57 it is about 1255.8, shooting up. Just past it, at 1.6, the tangent is about negative 34.23 - the sign flip you expect across a vertical asymptote.
Check
Two things restrict this function. Find both before you answer.
\[ f(x) = \frac{\sqrt{x-2}}{x-5} \]
Check your understanding
On which set is this function continuous?
Answer: A
Why: The square root requires the input to be at least two, and the denominator forbids five. At the input two the function is right-continuous, with value zero over negative three, so two belongs in the set. Everything else from two on, except five, is continuous.
Section
Part 4
Concept
Feeding one continuous function into another produces a continuous function. This is the last closure rule, and the most useful one.
\[ g \ \text{continuous at } a \ \text{ and } \ f \ \text{continuous at } g(a) \ \Longrightarrow \ f \circ g \ \text{continuous at } a \]
Read the second hypothesis carefully. The outer function has to be continuous at the output of the inner one, not at the original input.
Picture it
Animation
Shows: Composites of continuous functions are continuous — a rendered Manim animation.
Rendered with Manim.
Takeaway: This is what lets you build enormous continuous functions with no work.
Concept
Rewriting that closure rule as a limit statement gives the single most-used computational tool of the chapter.
\[ \lim_{x \to a} f\big(g(x)\big) = f\left(\lim_{x \to a} g(x)\right) \]
In words: if the outer function is continuous at the inner limit, you may move the limit symbol past it and evaluate the inside first.
The hypothesis is not decoration. The swap is legal only when the outer function is continuous at the value the inside is approaching.
Picture it
Animation
Shows: The limit slides inside a continuous function — a rendered Manim animation.
Rendered with Manim.
Takeaway: The swap is a theorem, not a notational convenience.
Intuition
Think of the outer function as a machine that never surprises you. Continuity means: inputs that are close together come out close together.
The inner function is delivering inputs that crowd in around some target value. A continuous machine turns that crowd of nearly-equal inputs into a crowd of nearly-equal outputs, all bunched around the machine's reading at the target.
If the machine has a jump right at the target, the crowd gets split in two - some inputs land just below the jump and some just above, and the outputs never agree. That is exactly when the swap fails.
Ranking
Put in order
Put the moves of Worked example: evaluate a limit by passing it inside into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. The outer is the square root and the inner is the rational expression.
Worked example
The inside is an indeterminate form, so simplify it first, then let the square root eat the limit.
\[ \lim_{x \to 2} \sqrt{\frac{x^2-4}{x-2}} \]
Name the outer and inner functions
Why: The outer is the square root and the inner is the rational expression. Passing the limit inside is only allowed once we know the square root is continuous at the inner limit.
\[ f(u) = \sqrt{u}, \qquad g(x) = \frac{x^2-4}{x-2} \]
Compute the inner limit by factoring and cancelling
Why: Direct substitution inside gives zero over zero, so factor. The shared factor cancels for every input except two, and the limit never inspects two itself.
\[ \lim_{x \to 2} \frac{(x-2)(x+2)}{x-2} = \lim_{x \to 2}(x+2) = 4 \]
Check the outer function at that value
Why: The square root is continuous on all nonnegative inputs, and four is comfortably positive. The hypothesis holds, so the swap is licensed.
\[ \sqrt{u} \ \text{is continuous at } u = 4 \]
Move the limit inside and evaluate
Why: With the outer function continuous at four, the limit of the root is the root of the limit.
\[ \lim_{x \to 2} \sqrt{\frac{x^2-4}{x-2}} = \sqrt{4} = 2 \]
Verify with values on both sides of two
Why: At input 2.01 the inside is 4.01 and its square root is about 2.0025. At input 1.99 the inside is 3.99 and its square root is about 1.9975. Both squeeze toward two from opposite sides, confirming the answer.
| input | inside value | square root |
|---|---|---|
| 1.99 | 3.99 | 1.99750 |
| 1.999 | 3.999 | 1.99975 |
| 2.001 | 4.001 | 2.00025 |
| 2.01 | 4.01 | 2.00250 |
Picture it
Animation
Shows: Each line of the worked example "evaluate a limit by passing it inside", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The square root is continuous on all nonnegative inputs, and four is comfortably positive. The hypothesis holds, so the swap is licensed.
Pattern
Predict first
The table runs: 0.99 | 0.9801 | first | 1.9602 · 0.999 | 0.998001 | first | 1.996002 · 1.001 | 1.002001 | second | 5.004002
In Trap: passing a limit inside a function that jumps, given the rows so far: what is the next one — the row where input is 1.01?
Correct: 1.01 | 1.0201 | second | 5.0402
| input | square | rule | output |
|---|---|---|---|
| 0.99 | 0.9801 | first | 1.9602 |
| 0.999 | 0.998001 | first | 1.996002 |
| 1.001 | 1.002001 | second | 5.004002 |
| 1.01 | 1.0201 | second | 5.0402 |
Why: The relationship between the columns, not the individual numbers, is what generates the next row. The inner limit really is one, so this first half is fine.
Trap
The student swaps the limit inside without checking the outer function is continuous there.
\[ h(u) = \begin{cases} 2u, & u < 1 \\ 2u+3, & u \ge 1 \end{cases}, \qquad \lim_{x \to 1} h(x^2) \]
Compute the inner limit, then plug it into the outer rule
Why: The inner limit really is one, so this first half is fine.
\[ \lim_{x \to 1} x^2 = 1 \ \Longrightarrow \ h(1) = 2(1)+3 = 5 \]
Report the limit as five
Why: Wrong. The outer function jumps at exactly the value the inside approaches, so the swap was never licensed. Values just left of one give roughly 1.96, not 5.
\[ \text{claimed: } \lim_{x \to 1} h(x^2) = 5 \quad (\text{false}) \]
Check the outer function at the inner limit before swapping, then fall back on one-sided limits.
\[ h(u) = \begin{cases} 2u, & u < 1 \\ 2u+3, & u \ge 1 \end{cases}, \qquad \lim_{x \to 1} h(x^2) \]
Test the outer function at the inner limit
Why: The outer function has a jump at one: the left limit there is two and the value is five. It is not continuous at one, so the limit may not be moved inside.
\[ \lim_{u \to 1^{-}} h(u) = 2 \ne 5 = h(1) \]
Do it honestly with one-sided limits
Why: Approaching one from below makes the square slightly less than one, so the first rule fires and the output nears two. Approaching from above makes the square slightly more than one, so the second rule fires and the output nears five.
\[ \lim_{x \to 1^{-}} h(x^2) = 2, \qquad \lim_{x \to 1^{+}} h(x^2) = 5 \]
Conclude the limit does not exist, and check it numerically
Why: At input 0.99 the square is 0.9801 and the output is 1.9602. At input 1.01 the square is 1.0201 and the output is 5.0402. The two sides disagree, so there is no limit at all - let alone five.
| input | square | rule | output |
|---|---|---|---|
| 0.99 | 0.9801 | first | 1.9602 |
| 0.999 | 0.998001 | first | 1.996002 |
| 1.001 | 1.002001 | second | 5.004002 |
| 1.01 | 1.0201 | second | 5.0402 |
Pattern
Step through it
Step through Trap: passing a limit inside a function that jumps one row at a time. What is driving the change, and what would the row after the last one be?
Pattern
Almost every routine limit you will ever compute is this one decision tree.
Continuity is what makes step one possible at all. Every time you have plugged a number into a formula to get a limit, you were quietly using it.
Check
Simplify the inside first, then apply the outer operation.
\[ \lim_{x \to 3} \left( \frac{x^2-9}{x-3} \right)^{2} \]
Check your understanding
What is the value of this limit?
Answer: A
Why: Inside, the numerator factors and the shared factor cancels, leaving x plus three, whose limit at three is six. Squaring is continuous everywhere, so the limit of the square is the square of the limit: six squared is thirty-six.
Section
Part 5
Concept
A piecewise function is usually built from ordinary formulas. On the inside of each piece it inherits the continuity of that formula for free.
The only inputs that need real work are the seams - the places where the rule changes hands - plus any point where an individual piece would break on its own.
seam — An input where a piecewise definition switches from one formula to another. The one-sided limits there come from two different rules, so they have to be compared by hand.
Matching
Match the pairs
Match each term to the definition this lesson gave it — not the one you would guess from the word.
Why: These are the working definitions of continuous at a point, one-sided continuity, seam as Continuity and the Intermediate Value Theorem uses them. Pairing them correctly is the test of whether you could state each one with the slide switched off.
Intuition
When a problem hands you an unknown constant inside one of the pieces, picture that constant as a dial that shifts or steepens that piece of the graph.
The two pieces currently end at different heights over the seam, so there is a visible gap. Turning the dial slides one end up or down.
Continuity is the setting where the gap closes exactly. That is one condition, so one unknown usually means one equation - and two seams with two unknowns means a system of two.
Step zero
Discussion prompt
Worked example: solve for the constant — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Note that only the seam at three is in question
Answer:
Worked example
Find the value of the constant that makes this function continuous everywhere.
\[ f(x) = \begin{cases} x^2 + k, & x \le 3 \\ 2x + 1, & x > 3 \end{cases} \]
Note that only the seam at three is in question
Why: Each piece is a polynomial, so it is continuous on its own stretch no matter what the constant is. The seam is the only place the two rules have to agree.
Compute the left-hand limit and the value at the seam
Why: The first rule covers inputs up to and including three, so it supplies both the left-hand limit and the function value there.
\[ \lim_{x \to 3^{-}} f(x) = 9 + k = f(3) \]
Compute the right-hand limit
Why: Inputs just above three use the second rule, which gives six plus one.
\[ \lim_{x \to 3^{+}} f(x) = 2(3) + 1 = 7 \]
Set the two sides equal and solve
Why: Continuity requires the two one-sided limits to agree with each other and with the value. Since the value already equals the left limit, one equation finishes it.
\[ 9 + k = 7 \ \Longrightarrow \ k = -2 \]
Verify by substituting the constant back in
Why: With the constant equal to negative two the first rule at three gives nine minus two, which is seven, and the second rule gives seven. Value, left limit, and right limit all equal seven, so all three conditions hold.
| quantity | expression | value |
|---|---|---|
| left limit | 9 + k | 7 |
| function value | 9 + k | 7 |
| right limit | 2(3) + 1 | 7 |
Picture it
Animation
Shows: Each line of the worked example "solve for the constant", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: With the constant equal to negative two the first rule at three gives nine minus two, which is seven, and the second rule gives seven. Value, left limit, and right limit all equal seven, so all three conditions hold.
Estimation
Predict first
Find both constants so that this function is continuous on the whole real line.
Commit before you compute: what does Worked example: two seams, two constants come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify both seams with the constants in place
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. At one: the left rule gives two plus two, which is four, and the middle rule gives four.
Worked example
Find both constants so that this function is continuous on the whole real line.
\[ f(x) = \begin{cases} 2x + a, & x < 1 \\ x^2 + 3, & 1 \le x \le 2 \\ bx - 1, & x > 2 \end{cases} \]
Identify both seams
Why: The rule changes at one and again at two. Each seam gives one continuity equation, and there are two unknowns, so the count matches.
Match the pieces at the first seam
Why: From the left the first rule gives two plus the constant. The value at one comes from the middle rule, which gives one plus three, or four. Setting them equal isolates the first constant immediately.
\[ 2(1) + a = 1^2 + 3 \ \Longrightarrow \ 2 + a = 4 \ \Longrightarrow \ a = 2 \]
Find the value at the second seam
Why: The middle rule covers the input two, so it supplies both the left-hand limit and the value there: four plus three.
\[ f(2) = 2^2 + 3 = 7 \]
Match the third piece to that height
Why: Inputs just above two use the third rule, which gives twice the second constant minus one. Setting that equal to seven pins the constant down.
\[ 2b - 1 = 7 \ \Longrightarrow \ 2b = 8 \ \Longrightarrow \ b = 4 \]
Verify both seams with the constants in place
Why: At one: the left rule gives two plus two, which is four, and the middle rule gives four. At two: the middle rule gives seven and the right rule gives eight minus one, which is seven. Both gaps are closed, so the function is continuous everywhere.
| seam | from the left | from the right | match? |
|---|---|---|---|
| 1 | 2(1) + 2 = 4 | 1 + 3 = 4 | yes |
| 2 | 4 + 3 = 7 | 4(2) - 1 = 7 | yes |
Picture it
Animation
Shows: Each line of the worked example "two seams, two constants", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At one: the left rule gives two plus two, which is four, and the middle rule gives four. At two: the middle rule gives seven and the right rule gives eight minus one, which is seven. Both gaps are closed, so the function is continuous everywhere.
Ranking
Put in order
These are the steps of Pattern: choosing parameters for continuity, scrambled. Put them back in order before the next slide shows you.
Why: This is the order the recipe itself gives. Recalling the sequence without the slide in front of you is the difference between recognising the method and being able to run it — most of what goes wrong in practice is a step done out of turn.
Pattern
Every problem of this shape runs the same five moves.
Count first: as many equations as seams, as many unknowns as constants. If those numbers do not match, expect either no solution or a whole family of them.
Edge cases
Discussion prompt
Pattern: choosing parameters for continuity works on the cases you have just seen. Push it to the edge: what is the most degenerate input it still handles — empty, zero, one item, everything equal — and what is the first case where it stops being true? Name the case, not just "it breaks".
Hint: Try the smallest legal input, then the largest, then the one where two things collide. Methods are specified at their edges; the middle takes care of itself.
Answer:
Every problem of this shape runs the same five moves.
Anomaly
Predict first
A student writes this, and it looks reasonable:
With two unknowns, the student closes the first gap, finds a pair that works there, and stops.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The left rule gives two at the input one, so the middle rule must too.
Two seams give two equations. Solve them together.
Why: The left rule gives two at the input one, so the middle rule must too. That is one equation in two unknowns - infinitely many pairs satisfy it.
Trap
With two unknowns, the student closes the first gap, finds a pair that works there, and stops.
\[ f(x) = \begin{cases} 3x-1, & x \le 1 \\ ax+b, & 1 < x < 4 \\ x^2-5, & x \ge 4 \end{cases} \]
Match at the first seam and pick any solution
Why: The left rule gives two at the input one, so the middle rule must too. That is one equation in two unknowns - infinitely many pairs satisfy it.
\[ a + b = 2 \quad \Rightarrow \quad a = 1,\ b = 1 \ (\text{a guess}) \]
Declare the function continuous
Why: Wrong. Nobody checked the second seam. With that guess the middle rule at four gives five while the last rule gives eleven - a jump of six sitting in plain sight.
\[ \lim_{x \to 4^{-}} f(x) = 5, \qquad f(4) = 16-5 = 11 \]
Two seams give two equations. Solve them together.
\[ f(x) = \begin{cases} 3x-1, & x \le 1 \\ ax+b, & 1 < x < 4 \\ x^2-5, & x \ge 4 \end{cases} \]
Write both seam equations before solving anything
Why: At one the left rule gives two; at four the right rule gives eleven. The middle rule has to hit both of those heights.
\[ a + b = 2, \qquad 4a + b = 11 \]
Subtract to eliminate the second unknown
Why: Subtracting the first equation from the second removes the constant term and leaves three times the slope equal to nine.
\[ 3a = 9 \ \Longrightarrow \ a = 3, \quad b = 2 - 3 = -1 \]
Verify both seams
Why: At one the middle rule gives three minus one, which is two, matching the left rule. At four it gives twelve minus one, which is eleven, matching sixteen minus five. Both gaps closed.
| seam | outer rule | middle rule | match? |
|---|---|---|---|
| 1 | 3(1) - 1 = 2 | 3(1) - 1 = 2 | yes |
| 4 | 16 - 5 = 11 | 3(4) - 1 = 11 | yes |
Trade off
Comparison matrix
From Trap: one equation for two seams: every row here is a choice with a cost. Fill the outer rule column, then say which row you would actually pick and what you give up for it.
| seam | outer rule | middle rule | match? |
|---|---|---|---|
| 1 | 3(1) - 1 = 2 | 3(1) - 1 = 2 | yes |
| 4 | 16 - 5 = 11 | 3(4) - 1 = 11 | yes |
Check
Match the two sides at the seam, then solve for the unknown.
\[ f(x) = \begin{cases} c x^2, & x < 2 \\ x + 6, & x \ge 2 \end{cases} \]
Check your understanding
Which value of the constant makes this function continuous at the input two?
Answer: A
Why: The left-hand limit is the constant times two squared, which is four times the constant. The value and the right-hand limit are two plus six, which is eight. Setting four times the constant equal to eight gives the constant equal to two.
Section
Part 6
Intuition
Suppose you take the stairs from the basement to the third floor. At some moment you were on the second floor. Not because anyone measured it - because there is no other way to get there.
That is the whole Intermediate Value Theorem. An unbroken path from a low value to a high value must pass through every value in between.
Notice what the theorem does not tell you: which step, or how many times. It only promises the crossing happened somewhere. Existence, not location.
And notice why the stairs mattered. In an elevator with a broken shaft you could teleport past floor two - so the unbroken part is doing all the work.
Picture it
Figure (svg): A rising curve from a marked left endpoint to a marked right endpoint, crossing a dashed horizontal target line at one circled point in between
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
Here is the formal statement. Read the hypotheses slowly - they are what students drop.
Concept
Here is the formal statement. Read the hypotheses slowly - they are what students drop.
\[ \text{If } f \text{ is continuous on } [a,b] \text{ and } N \text{ is between } f(a) \text{ and } f(b), \]
\[ \text{then there exists } c \in (a,b) \text{ with } f(c) = N. \]
Figure (svg): A rising curve from a marked left endpoint to a marked right endpoint, crossing a dashed horizontal target line at one circled point in between
The conclusion is an existence statement, and the guaranteed point is strictly inside the interval, never at an endpoint.
Explain it
Discussion prompt
Explain The Intermediate Value Theorem to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
The conclusion is an existence statement, and the guaranteed point is strictly inside the interval, never at an endpoint.
Picture it
Animation
Shows: A continuous cubic passing from negative to positive values.
It crossed. Where, the theorem will not say.
Takeaway: Negative at one end, positive at the other, and no breaks in between — so somewhere it must have passed through zero. The theorem promises existence, not a location.
Concept
Two things must be checked before you may cite the theorem, and a written solution that skips them is incomplete even if the answer is right.
Drop the first and the conclusion genuinely fails - you will see two counterexamples in a moment. Drop the second and you are simply citing a theorem whose conditions were never met.
In practice the continuity check is one sentence: name the family the function belongs to and note the interval sits inside its domain.
Analogy
Discussion prompt
Explain The hypotheses are doing real work by analogy to something with no Calculus I in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Two things must be checked before you may cite the theorem, and a written solution that skips them is incomplete even if the answer is right.
Picture it
Animation
Shows: A jump discontinuity stepping over the value zero.
A jump skips the floor entirely.
Takeaway: A function that jumps can step straight over the value you were promised. Continuity is what forbids that leap, which is why the hypothesis cannot be dropped.
Concept
The special case you will use most often is the one where the target height is zero.
\[ f \text{ continuous on } [a,b] \ \text{ and } \ f(a)\,f(b) < 0 \ \Longrightarrow \ f(c) = 0 \ \text{ for some } c \in (a,b) \]
A negative product just says the two endpoint values have opposite signs. Between a value below the axis and a value above it, an unbroken curve must touch the axis.
To show an equation has a solution, move everything to one side so the question becomes a root question, then hunt for a sign change.
Counterexample
Discussion prompt
The special case you will use most often is the one where the target height is zero.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
To show an equation has a solution, move everything to one side so the question becomes a root question, then hunt for a sign change.
Estimation
Predict first
Prove that this equation has a solution somewhere between one and two.
Commit before you compute: what does Worked example: show a cubic has a root come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify by trapping the root more tightly
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. At 1.3 the value is 2.197 minus 2.3, which is negative 0.103.
Worked example
Prove that this equation has a solution somewhere between one and two.
\[ x^3 - x - 1 = 0 \]
Name the function and state its continuity
Why: The left side is a polynomial, and polynomials are continuous on every closed interval. That is hypothesis one, checked and written down.
\[ f(x) = x^3 - x - 1 \ \text{ is continuous on } [1,2] \]
Evaluate at the left endpoint
Why: One cubed is one, minus one is zero, minus one more is negative one. The curve starts below the axis.
\[ f(1) = 1 - 1 - 1 = -1 \]
Evaluate at the right endpoint
Why: Two cubed is eight, minus two is six, minus one is five. The curve ends above the axis, so a sign change has occurred.
\[ f(2) = 8 - 2 - 1 = 5 \]
Cite the theorem and state the conclusion
Why: Zero lies strictly between negative one and five, and the function is continuous on the closed interval, so the theorem hands over a point inside where the value is zero.
\[ \exists\, c \in (1,2) \ \text{ with } \ f(c) = 0 \]
Verify by trapping the root more tightly
Why: At 1.3 the value is 2.197 minus 2.3, which is negative 0.103. At 1.4 it is 2.744 minus 2.4, which is positive 0.344. The sign flips between them, so the root really does live near 1.32 - inside the interval as promised.
| input | value of f | sign |
|---|---|---|
| 1 | -1 | negative |
| 1.3 | -0.103 | negative |
| 1.4 | 0.344 | positive |
| 2 | 5 | positive |
Picture it
Animation
Shows: Each line of the worked example "show a cubic has a root", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At 1.3 the value is 2.197 minus 2.3, which is negative 0.103. At 1.4 it is 2.744 minus 2.4, which is positive 0.344. The sign flips between them, so the root really does live near 1.32 - inside the interval as promised.
Step zero
Discussion prompt
Worked example: show an equation has a solution — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Move everything to one side
Answer:
Worked example
This equation cannot be solved with algebra, but the theorem still proves a solution exists.
\[ \cos x = x \]
Move everything to one side
Why: The theorem's root form needs a single function set equal to zero, so subtract the right side from the left.
\[ g(x) = \cos x - x \]
State continuity on the chosen interval
Why: Cosine is continuous everywhere and so is the identity function, and a difference of continuous functions is continuous. Any closed interval will do; take the one from zero to one.
\[ g \ \text{ is continuous on } [0,1] \]
Evaluate at the endpoints, in radians
Why: At zero, cosine is one and the input is zero, giving one. At one radian, cosine is about 0.5403, so the difference is about negative 0.4597. Opposite signs.
\[ g(0) = 1 > 0, \qquad g(1) \approx -0.4597 < 0 \]
Apply the theorem
Why: Zero lies between a positive and a negative endpoint value, and the function is continuous on the closed interval, so some interior point sends it to zero. At that point cosine equals the input.
\[ \exists\, c \in (0,1) \ \text{ with } \ \cos c = c \]
Verify by narrowing to two decimal places
Why: At 0.7 the value is 0.7648 minus 0.7, which is positive 0.0648. At 0.8 it is 0.6967 minus 0.8, which is negative 0.1033. The sign change traps the solution between 0.7 and 0.8, and it is about 0.7391.
| input (radians) | cosine | value of g | sign |
|---|---|---|---|
| 0 | 1 | 1 | positive |
| 0.7 | 0.76484 | 0.06484 | positive |
| 0.8 | 0.69671 | -0.10329 | negative |
| 1 | 0.54030 | -0.45970 | negative |
Picture it
Animation
Shows: Each line of the worked example "show an equation has a solution", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At 0.7 the value is 0.7648 minus 0.7, which is positive 0.0648. At 0.8 it is 0.6967 minus 0.8, which is negative 0.1033. The sign change traps the solution between 0.7 and 0.8, and it is about 0.7391.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The endpoint values have opposite signs, so the student declares a root and moves on.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Both computations are correct: the function really is negative one on the left and positive one on the right.
Check continuity on the closed interval first, before looking at any signs.
Why: Both computations are correct: the function really is negative one on the left and positive one on the right.
Trap
The endpoint values have opposite signs, so the student declares a root and moves on.
\[ f(x) = \frac{1}{x} \quad \text{on} \quad [-1,\,1] \]
Evaluate the endpoints and spot the sign change
Why: Both computations are correct: the function really is negative one on the left and positive one on the right.
\[ f(-1) = -1 < 0, \qquad f(1) = 1 > 0 \]
Conclude some interior point gives zero
Why: Wrong. The continuity hypothesis was never checked, and it fails badly - the function is not even defined at zero. There is no such point, because this function is never zero for any input at all.
\[ \frac{1}{c} = 0 \ \text{ has no solution} \]
Check continuity on the closed interval first, before looking at any signs.
\[ f(x) = \frac{1}{x} \quad \text{on} \quad [-1,\,1] \]
Test the hypothesis: is the function continuous on the whole closed interval?
Why: No. The input zero lies inside the interval and the function has an infinite discontinuity there, so the theorem does not apply and nothing may be concluded.
\[ \lim_{x \to 0^{-}} \frac{1}{x} = -\infty, \qquad \lim_{x \to 0^{+}} \frac{1}{x} = +\infty \]
See what the graph actually does
Why: The sign flip happens by jumping through the asymptote, not by crossing the axis. That is exactly the teleporting elevator the theorem was built to rule out.
| input | value | note |
|---|---|---|
| -0.1 | -10 | diving down |
| -0.01 | -100 | no crossing here |
| 0.01 | 100 | reappears on top |
| 0.1 | 10 | still never zero |
Comparison
Comparison matrix
From Trap: using the theorem across a vertical asymptote: refill the note column from what you know. The rest of the table is as it appeared.
| input | value | note |
|---|---|---|
| -0.1 | -10 | diving down |
| -0.01 | -100 | no crossing here |
| 0.01 | 100 | reappears on top |
| 0.1 | 10 | still never zero |
Anomaly
Predict first
A student writes this, and it looks reasonable:
The student finds a sign change and reports a unique root.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: At negative three the value is negative twenty-seven plus twelve, or negative fifteen.
Report exactly what the theorem promises - at least one - and factor if you actually want the count.
Why: At negative three the value is negative twenty-seven plus twelve, or negative fifteen. At three it is twenty-seven minus twelve, or fifteen. A genuine sign change, and the function is a polynomial, so the theorem does apply.
Trap
The student finds a sign change and reports a unique root.
\[ f(x) = x^3 - 4x \quad \text{on} \quad [-3,\,3] \]
Check the endpoints and apply the theorem
Why: At negative three the value is negative twenty-seven plus twelve, or negative fifteen. At three it is twenty-seven minus twelve, or fifteen. A genuine sign change, and the function is a polynomial, so the theorem does apply.
\[ f(-3) = -15, \qquad f(3) = 15 \]
Conclude there is exactly one root in the interval
Why: Wrong. The theorem's conclusion is an existence claim with no uniqueness attached. Nothing in its statement counts crossings.
\[ \text{claimed: exactly one } c \ \text{ with } f(c) = 0 \]
Report exactly what the theorem promises - at least one - and factor if you actually want the count.
\[ f(x) = x^3 - 4x \quad \text{on} \quad [-3,\,3] \]
State the correct conclusion
Why: The sign change guarantees the curve meets the axis somewhere inside. That is the entire content of the theorem.
\[ \exists\, c \in (-3,3) \ \text{ with } \ f(c) = 0 \]
Factor to see how many roots there really are
Why: This cubic factors completely, and it has three roots inside the interval, not one. Uniqueness needs a separate argument - usually that the derivative never changes sign.
\[ x^3 - 4x = x(x-2)(x+2) \ \Longrightarrow \ x = -2,\,0,\,2 \]
Verify the three roots
Why: Substituting negative two gives negative eight plus eight, which is zero. Zero gives zero. Two gives eight minus eight, which is zero. Three crossings inside an interval where the theorem promised only one.
| input | value of f |
|---|---|
| -2 | 0 |
| 0 | 0 |
| 2 | 0 |
Invariant
Step through it
Step through Trap: "at least one" is not "exactly one" one row at a time. One of these columns never changes — find it, and say why it cannot.
Pattern
Graders want these four sentences, in this order. Missing any one costs points even when the answer is right.
If the endpoints do not straddle the target, do not force it - pick a different interval, or check a few values to find where the sign actually changes.
Real world
Discussion prompt
Outside this lesson: where does Continuity and the Intermediate Value Theorem actually turn up? Name one concrete situation — a job, a piece of software someone ships, a decision somebody has to make — and say which part of Pattern: the complete root-existence argument is doing the work in it.
Hint: Vague is the failure mode here. "Engineering" is not a situation; "deciding whether this build is fast enough to ship" is.
Answer:
That deck gives the three-part definition of continuity at a point and uses it to classify removable, jump, and infinite discontinuities. It then covers continuity on intervals and for the standard function families, composites and passing a limit inside, and choosing parameters that make a piecewise function continuous, before closing with the Intermediate Value Theorem and bisection. It targets the students who check the limit but forget that the function must be defined, who think cancelling a factor erases the hole, and who use the IVT without checking continuity or read it as guaranteeing exactly one root.
Concept
Existence is nice, but sooner or later you want the root itself. Bisection is the theorem applied over and over.
\[ \text{after } n \text{ steps the root is trapped in an interval of width } \frac{b-a}{2^{n}} \]
It is slow but it never fails: every step is another application of the theorem, so the root can never escape the interval you are holding.
Picture it
Animation
Shows: Bisection turns existence into a number — a rendered Manim animation.
Rendered with Manim.
Takeaway: The IVT promises a root. Bisection goes and finds it.
Ranking
Put in order
Put the moves of Worked example: three rounds of bisection into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. The midpoint is 1.5, where the cube is 3.375, so the value is 3.375 minus 2.5, or positive 0.875.
Worked example
Narrow the root of the cubic from earlier, starting from the interval that the theorem already gave us.
\[ f(x) = x^3 - x - 1 \quad \text{on} \quad [1,\,2] \]
Round 1: test the midpoint of the starting interval
Why: The midpoint is 1.5, where the cube is 3.375, so the value is 3.375 minus 2.5, or positive 0.875. The left endpoint was negative, so the sign change lives in the left half.
\[ f(1.5) = 0.875 > 0 \ \Longrightarrow \ \text{root in } [1,\,1.5] \]
Round 2: bisect again
Why: The new midpoint is 1.25, where the cube is 1.953125, so the value is 1.953125 minus 2.25, or negative 0.296875. Now the negative end is 1.25 and the positive end is 1.5.
\[ f(1.25) = -0.296875 < 0 \ \Longrightarrow \ \text{root in } [1.25,\,1.5] \]
Round 3: bisect once more
Why: The midpoint is 1.375, where the cube is 2.599609375, so the value is that minus 2.375, or positive 0.224609375. The sign change is now between 1.25 and 1.375.
\[ f(1.375) = 0.224609375 > 0 \ \Longrightarrow \ \text{root in } [1.25,\,1.375] \]
Report the estimate and its guaranteed error
Why: Three halvings shrank the interval from a width of one to a width of one eighth. Taking the midpoint of the final interval gives an estimate whose error is at most one sixteenth.
\[ c \approx 1.3125 \quad \text{with error at most } 0.0625 \]
Verify against the true root
Why: The actual root is about 1.32472. That sits inside the final interval from 1.25 to 1.375, and it is 0.0122 away from the estimate 1.3125 - comfortably inside the promised bound of 0.0625.
| round | midpoint | value of f | interval kept |
|---|---|---|---|
| start | - | - | [1, 2] |
| 1 | 1.5 | 0.875 | [1, 1.5] |
| 2 | 1.25 | -0.296875 | [1.25, 1.5] |
| 3 | 1.375 | 0.224609375 | [1.25, 1.375] |
Picture it
Animation
Shows: Each line of the worked example "three rounds of bisection", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The actual root is about 1.32472. That sits inside the final interval from 1.25 to 1.375, and it is 0.0122 away from the estimate 1.3125 - comfortably inside the promised bound of 0.0625.
Elimination
Eliminate the wrong options
What does the Intermediate Value Theorem let you conclude on the interval from zero to one?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: The function is a polynomial, hence continuous on the closed interval, and the endpoint values one and negative two have opposite signs, so zero lies between them. The theorem then guarantees at least one interior input where the value is zero.
Check
A polynomial has been evaluated at both endpoints of an interval. Decide what may honestly be concluded.
\[ f(x) = x^3 - 4x + 1, \qquad f(0) = 1, \qquad f(1) = -2 \]
Check your understanding
What does the Intermediate Value Theorem let you conclude on the interval from zero to one?
Answer: A
Why: The function is a polynomial, hence continuous on the closed interval, and the endpoint values one and negative two have opposite signs, so zero lies between them. The theorem then guarantees at least one interior input where the value is zero.
Check
A student writes the argument below. Find the flaw.
\[ f(x) = \frac{1}{x-2} \ \text{ on } [1,3]: \quad f(1) = -1,\ f(3) = 1 \ \Rightarrow \ f(c) = 0 \ \text{ for some } c \]
Check your understanding
What is wrong with this argument?
Answer: A
Why: The input two lies inside the interval and makes the denominator zero, so the function is undefined there and blows up on both sides. The continuity hypothesis fails, and indeed this function is never zero for any input, so the conclusion is false.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — Continuity at a Point · Classifying Discontinuities · Continuity on an Interval · Composites and Passing the Limit Inside · Making a Piecewise Function Continuous · The Intermediate Value Theorem. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
Continuity is the promise that a function has no surprises: the value at a point is exactly what its neighbors predict.
| When you see | Do this |
|---|---|
| zero over zero at a point | factor and cancel, then compare the limit with the value |
| a piecewise seam | compute both one-sided limits and the value there |
| a denominator that vanishes and does not cancel | infinite discontinuity, vertical asymptote |
| an unknown constant in a piece | set left limit equal to right limit and solve |
| prove a solution exists | define the function, assert continuity, show a sign change, cite the theorem |
Next up: limits at infinity and end behavior - the same limit machinery, but pointed at what happens far out on the axis instead of at a single point.
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