A function is continuous where you can draw it without lifting your pen. That informal test is exactly what the syllabus asks for, and it is enough to classify every break you will meet.
Today's hook, You do not need limits to decide whether a graph has a break in it. The syllabus defines continuity by whether the pen leaves the paper, and that definition does all the work at this level.
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Recall, your gut answer first
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
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Recall, your gut answer first
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Sketch $y = \dfrac{1}{x}$. Could you draw the whole thing without lifting your pen? Now sketch a function that costs $\$3$ for up to 2 kg and $\$5$ above that. Same question.
Before you work it out, what is your instinct? Write it down, then check it against the lesson.
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Draw it without lifting the pen
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Draw it without lifting the pen
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A function is continuous at a point if the curve passes through that point without any break, so you could draw it there without lifting your pen. A discontinuity is simply a point where that fails. Everything else in this lesson is classifying the ways it can fail.
Continuous at $a$: no break at $x = a$ Discontinuity at $a$: the pen must lift at $x = a$
It is a local property
Continuity is decided point by point. $y = \frac{1}{x}$ is continuous at every point of its domain and discontinuous only at $x = 0$, which is not in its domain at all.
Piecewise breaks happen at the joins
Each piece is usually continuous on its own. Check only where the definition changes over.
Three ways to break
A jump, a hole, or a vertical asymptote. Naming which one is what the identification dot-point asks for.
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What you'll master
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
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What you'll master
Know
Key facts
A function is continuous at a point if the curve can be drawn through it without lifting the pen.
A discontinuity is a point where the function is not continuous.
The three types you meet are a jump, a hole (removable), and an infinite discontinuity at a vertical asymptote.
For a piecewise function, check continuity at each join by comparing the two pieces there.
Understand
Concepts
Why continuity is decided at a point rather than for the whole function at once.
Why a piecewise function can only break where its definition changes.
Why a hole and a jump are different failures even though both need the pen lifted.
Can do
Skills
State whether a function is continuous at a given point and justify it informally.
Identify and classify every discontinuity of a given function or graph.
Test a piecewise function for continuity at its joins.
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Key terms
Continuous at a pointNo break in the curve there, so you could draw through it without lifting your pen. Like this: $y = x^2$ is continuous at $x = 3$ because the parabola passes smoothly through $(3, 9)$.
DiscontinuityA point where the function is not continuous, so the pen must lift. Like this: $y = \frac{1}{x}$ has a discontinuity at $x = 0$.
Jump discontinuityA break where the graph stops at one height and restarts at another. Like this: a parking fee of $\$5$ up to 2 hours and $\$9$ after jumps at 2 hours.
Removable discontinuityA single missing point, a hole, with the curve otherwise intact. Like this: $y = \frac{x^2-4}{x-2}$ is the line $y = x + 2$ with a hole at $x = 2$.
Infinite discontinuityA break where the curve shoots off towards a vertical asymptote. Like this: $y = \frac{1}{x}$ near $x = 0$.
Piecewise joinThe $x$-value where a piecewise function switches from one rule to the next, and the only place it can break. Like this: a function defined as $x + 1$ for $x < 2$ and $3x$ for $x \geq 2$ can only break at $x = 2$.
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The informal definition, and why it is enough
Work through the core explanation before applying it.
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The informal definition, and why it is enough
core concept
The syllabus defines continuity informally: a function is continuous at a point if the curve can be drawn through that point without lifting the pen off the paper. A discontinuity is a point where that is impossible.
That is a genuine mathematical test at this level. You do not need limits, and the Advanced course does not ask for them here. Reading the graph, or reasoning about the rule, is the expected method.
Continuity is a property of a **point**, not of the whole function. $y = \frac{1}{x}$ is continuous everywhere it is defined; it fails only at $x = 0$, which is not in its domain.
Most familiar functions are continuous everywhere they are defined. Polynomials, sine and cosine, and exponentials never break. So the question is always where the exceptions are: denominators that vanish, piecewise joins, and cancelled factors.
Quick check: at which value of $x$ is $y = \dfrac{1}{x - 3}$ discontinuous?
A function is continuous at a point if you can draw through it without lifting the pen; a discontinuity is where you cannot. Continuity is decided point by point. Polynomials, sine, cosine and exponentials are continuous everywhere; look for breaks at zero denominators, piecewise joins and cancelled factors.
Pause, copy the informal definition of continuity and discontinuity, the note that it is a point-by-point property, and the three places to look for breaks, into your book.
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The three ways a function breaks
core concept
We just saw that a discontinuity is a point where the pen must lift. That raises a question: is one break the same as another? This card answers it → no, and naming the type is what the identification dot-point asks for.
**A jump.** The graph stops at one height and restarts at another. Step-pricing models do this: a parking fee of $\$5$ up to 2 hours and $\$9$ after jumps at 2 hours.
**A hole, also called removable.** One single point is missing and everything else is intact. $y = \dfrac{x^2 - 4}{x - 2}$ simplifies to $y = x + 2$ everywhere except $x = 2$, where the original is undefined. The graph is the line with one point punched out.
**An infinite discontinuity.** The curve shoots away towards a vertical asymptote, as $y = \dfrac{1}{x}$ does at $x = 0$.
A hole comes from a cancelled factor. If the numerator and denominator share a factor, cancelling gives the shape of the graph but hides the hole. The hole sits at the value that made the cancelled factor zero, and it must still be marked.
Which is NOT a type of discontinuity you would meet in this course?
Three types: a jump (graph restarts at a different height), a hole or removable discontinuity (one point missing, usually from a cancelled factor), and an infinite discontinuity (the curve runs off to a vertical asymptote). Naming the type is part of the answer.
Pause, copy the three types with one example each, and the note that a cancelled factor leaves a hole at the value making that factor zero, into your book.
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Testing a piecewise function at its joins
core concept
We just saw how to classify a break once you have found it. That raises a question: for a piecewise function, where do you even look? This card answers it → only at the joins, because each piece is normally continuous on its own.
A piecewise function is built from ordinary rules, each continuous where it applies. So it can only break where the definition changes over, at the join.
To test a join at $x = c$: work out the value the first piece is heading to at $c$, and the value the second piece gives at $c$. If they agree, the pieces meet and the function is continuous there. If not, there is a jump.
For $f(x) = x + 1$ when $x < 2$ and $f(x) = 3x$ when $x \geq 2$: the first piece heads to $2 + 1 = 3$, the second gives $3(2) = 6$. They disagree, so there is a jump discontinuity at $x = 2$ of size 3.
Sketch the join. Marking an open circle where a piece stops and a filled circle where the next begins makes the answer visible, and it is the same open and closed circle convention you used for inequalities on a number line.
Fill the blank: for $f(x) = x + 1$ when $x < 4$ and $f(x) = 2x$ when $x \geq 4$, the first piece heads to 5 and the second gives 8, so there is a jump of size .
A piecewise function can only break at a join. Compare the value the earlier piece heads to with the value the later piece gives there: agreeing means continuous, disagreeing means a jump. Sketch the join with open and closed circles.
Pause, copy the join test, the worked $x+1$ against $3x$ at $x = 2$ giving a jump of 3, and the open and closed circle convention, into your book.
Worked examples · 3 in a row, reveal as you go
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Work examples end to end
Follow the reasoning through complete worked solutions.
PROBLEM 1 · CLASSIFYING A DISCONTINUITY
State where $y = \dfrac{x^2 - 9}{x - 3}$ is discontinuous, and what type of discontinuity it is.
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The denominator is zero at $x = 3$, so the function is undefined there
A function cannot be continuous at a point where it is not defined.
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Factorise: $\dfrac{(x-3)(x+3)}{x-3} = x + 3$ for $x \neq 3$
The factor cancels, so away from $x = 3$ this is just a straight line.
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The graph is the line $y = x + 3$ with a hole at $(3, 6)$, so it is a removable discontinuity
A cancelled factor leaves a hole, not an asymptote.
PROBLEM 2 · PIECEWISE JOIN
Determine whether $f(x) = \begin{cases} x + 1 & x < 2 \\ 3x & x \geq 2 \end{cases}$ is continuous at $x = 2$.
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First piece as $x$ approaches 2: $2 + 1 = 3$
The value the earlier rule is heading to.
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Second piece at $x = 2$: $3(2) = 6$
The value the later rule actually gives.
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$3 \neq 6$, so there is a jump discontinuity at $x = 2$
The pieces do not meet, so the pen must lift.
PROBLEM 3 · INFINITE DISCONTINUITY
Identify and classify every discontinuity of $y = \dfrac{2}{x^2 - 4}$.
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$x^2 - 4 = 0$ when $x = 2$ or $x = -2$
Find where the denominator vanishes.
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Neither factor cancels, since the numerator is the constant $2$
No cancellation means no hole.
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Infinite discontinuities at $x = 2$ and $x = -2$, with vertical asymptotes there
The curve runs off to infinity at both.
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Quick-fire practice
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Quick-fire practice
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Where is $y = \dfrac{1}{x + 5}$ discontinuous, and of what type?
Is $y = x^3 - 2x$ continuous everywhere? Justify in one line.
Classify the discontinuity of $y = \dfrac{x^2 - 1}{x - 1}$.
For $f(x) = 2x$ when $x < 1$ and $f(x) = x + 1$ when $x \geq 1$, is $f$ continuous at $x = 1$?
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Revisit your two sketches
Run the quick drill and copy the summary into your book.
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Revisit your two sketches
At the start you sketched $y = \frac{1}{x}$ and a step-pricing function. Name the type of discontinuity in each, and explain why the pricing one is unavoidable rather than a flaw in the model.
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Multiple choice
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Multiple choice
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Short answer
Write full responses, then check them against the model answers.
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Short answer
ApplyBand 43 marks
Q1. Identify every discontinuity of $y = \dfrac{x - 2}{x^2 - 4}$ and classify each one. (3 marks)
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ApplyBand 54 marks
Q2. The function $f(x) = \begin{cases} x^2 & x < 3 \\ ax + 1 & x \geq 3 \end{cases}$ is continuous at $x = 3$. Find the value of $a$, showing your reasoning. (4 marks)
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UnderstandBand 32 marks
Q3. State the informal definition of continuity used in this course, and explain why $y = \dfrac{1}{x}$ is described as continuous on its domain despite having a break at $x = 0$. (2 marks)
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📖 Comprehensive answers (click to reveal)
Practice 1: at $x = -5$, an infinite discontinuity (vertical asymptote). Practice 2: yes, it is a polynomial and polynomials are continuous everywhere. Practice 3: $\frac{(x-1)(x+1)}{x-1} = x+1$ with a hole at $(1,2)$, so removable. Practice 4: first piece heads to 2, second gives 2, so they agree and $f$ is continuous at $x = 1$.
Q1 (3 marks): $x^2 - 4 = (x-2)(x+2)$, so the function is undefined at $x = 2$ and $x = -2$ [1]. The factor $(x-2)$ cancels, leaving $\frac{1}{x+2}$, so $x = 2$ is a removable discontinuity, a hole at $(2, \frac{1}{4})$ [1]. $(x+2)$ does not cancel, so $x = -2$ is an infinite discontinuity with a vertical asymptote [1].
Q2 (4 marks): For continuity the two pieces must give the same value at $x = 3$ [1]. First piece: $3^2 = 9$ [1]. Second piece: $3a + 1$ [1]. Setting $3a + 1 = 9$ gives $a = \frac{8}{3}$ [1].
Q3 (2 marks): A function is continuous at a point if the curve can be drawn through that point without lifting the pen off the paper, and a discontinuity is a point where it cannot [1]. $x = 0$ is not in the domain of $\frac{1}{x}$, and continuity is only ever claimed at points where the function is defined, so the function is continuous at every point of its domain [1].
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Boss battle · Break Detector
earn bronze · silver · gold
Spot and classify discontinuities at speed, including piecewise joins. Beat the boss to bank a tier, gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.