What Is A Discontinuous Function?
A discontinuous function is a function that is not continuous at one or more points in its domain — meaning its graph has at least one break, hole, jump, or vertical asymptote. Where a continuous function can be traced in one unbroken stroke, a discontinuous function forces the pen off the paper at least once.
Formally, a function $f$ is continuous at $x = a$ when three conditions all hold: $f(a)$ exists, the limit $\lim_{x \to a} f(x)$ exists, and those two values are equal. If any one of the three fails, $f$ is discontinuous at $x = a$.
Symbol | Meaning |
|---|---|
$f(a)$ | the function's value at the point $a$ |
$\lim_{x \to a^-} f(x)$ | the left-hand limit (approaching $a$ from below) |
$\lim_{x \to a^+} f(x)$ | the right-hand limit (approaching $a$ from above) |
$\lim_{x \to a} f(x)$ | the two-sided limit (exists only when left and right agree) |
What Are the Types of Discontinuity?
There are three core types. Reading them off a graph is faster than testing limits algebraically, so learn the visual signature of each first.
Removable discontinuity (a hole). Both one-sided limits exist and agree, but the function either has no value there or a value sitting off the curve. The two-sided limit exists; the function value does not match it. You could "fix" it by redefining one point.
Jump discontinuity (a step). Both one-sided limits exist but are different. The graph leaps from one height to another, as the shipping rate did. The two-sided limit does not exist.
Infinite discontinuity (an asymptote). At least one one-sided limit runs off to $+\infty$ or $-\infty$. The graph shoots up or down a vertical asymptote, as in $f(x) = \frac{1}{x}$ at $x = 0$.
A fourth label, an oscillating discontinuity (such as $\sin\frac{1}{x}$ near $0$), shows up in higher courses. For most algebra and early-calculus work, the three above cover what you will meet.
How Do You Tell If a Function Is Discontinuous?
Run the three-part continuity test at any suspicious point — usually where a denominator hits zero, where a piecewise rule switches, or where an absolute value or floor changes behaviour.
Check whether $f(a)$ is defined.
Find the left-hand and right-hand limits.
Compare. If the limits disagree, it is a jump. If they agree but $f(a)$ differs or is missing, it is removable. If a limit is infinite, it is infinite.
A common reader question — "is a piecewise function always discontinuous?" — has a clean answer: no. A piecewise function is discontinuous only where the pieces fail to meet. If the pieces join at the same height, the function stays continuous there.
Examples of Discontinuous Functions
These six progress from a single clean break to mixed-sign, fraction-heavy, and real-world phrasing. Re-trace each on paper.
Example 1
Where is $f(x) = \dfrac{x - 2}{x - 4}$ discontinuous?
Set the denominator to zero.
$x - 4 = 0$
$x = 4$
At $x = 4$ the function is undefined and the one-sided limits run to $\pm\infty$.
Final answer: infinite discontinuity at $x = 4$.
Example 2
A student claims $g(x) = \dfrac{x^2 - 9}{x - 3}$ has an asymptote at $x = 3$. Is that right?
Wrong attempt. The denominator is zero at $x = 3$, so the student writes "vertical asymptote, infinite discontinuity" and stops. That mirrors Example 1, so it feels safe.
Why it breaks. Plug a value close to $3$, say $x = 2.99$: the output is near $5.99$, not blowing up. An asymptote would send the value toward infinity. Something is off.
Correct. Factor first.
$g(x) = \dfrac{(x-3)(x+3)}{x - 3}$
$g(x) = x + 3 \quad (x \neq 3)$
The factor cancels, leaving a hole at $x = 3$ where the curve would sit at $6$.
Final answer: removable discontinuity at $x = 3$, not infinite. Cancelling factors changes the type entirely.
Example 3
Find the discontinuity of the piecewise function $$h(x) = \begin{cases} x + 1, & x < 1 \ 4, & x = 1 \ x + 1, & x > 1 \end{cases}$$
The limit from both sides is $\lim_{x \to 1}(x+1) = 2$, but $h(1) = 4$.
The limit exists; the value does not match it.
Final answer: removable discontinuity at $x = 1$.
Example 4
Classify the discontinuity of $$p(x) = \begin{cases} x^2, & x \leq 2 \ x + 5, & x > 2 \end{cases}$$
Left-hand limit: $\lim_{x \to 2^-} x^2 = 4$.
Right-hand limit: $\lim_{x \to 2^+} (x + 5) = 7$.
The one-sided limits exist but differ ($4 \neq 7$).
Final answer: jump discontinuity at $x = 2$.
Example 5
Where is the greatest-integer (floor) function $f(x) = \lfloor x \rfloor$ discontinuous?
The floor function returns the largest integer less than or equal to $x$. At every integer it steps up by one.
At $x = 3$: left-hand limit is $2$, right-hand limit is $3$.
Final answer: jump discontinuity at every integer. (For the full behaviour, see the greatest integer function.)
Example 6
A parking garage charges $4 for the first hour or part of it, then $4 more for each additional hour or part. Where is the cost function discontinuous, and why does it matter?
The cost holds flat across each hour, then jumps $4 at every whole hour.
$x$ just under $1$ hour costs $4$; $x$ just over $1$ hour costs $8$.
Final answer: jump discontinuities at $x = 1, 2, 3, \dots$, and a driver who returns at 60 minutes and one second pays for a full extra hour.
[INTERACTIVE: A Desmos graph where the reader drags a point along a piecewise function and a panel reports the left-hand limit, right-hand limit, and function value at that x, flagging the discontinuity type live.]
Why Discontinuous Functions Matter
"Many real quantities change in steps, not smoothly."
Smoothness is the exception in the real world, not the rule. The reason the discontinuous function earns its own name is that the things we most need to model — prices, signals, thresholds — refuse to be smooth.
Pricing and tax. Postage, parking, and income-tax brackets all jump at thresholds. The function has to jump because the rule does.
Digital signals. A signal that flips between $0$ and $1$ is discontinuous by design — the jump is the information.
Calculus theorems. The Intermediate Value Theorem and the Fundamental Theorem of Calculus require continuity on an interval. Knowing where a function breaks tells you exactly where those tools stop working — which is why zeros of a function on a discontinuous piece need extra care.
The destination here is calculus: continuity is the gate every limit and derivative argument has to pass through. Spotting a break early saves you from applying a theorem where it does not hold.
Tripping Points To Avoid
Mistake 1: Calling every zero denominator an asymptote
Where it slips in: Any rational function where a factor cancels, as in Example 2.
Don't do this: Declare "denominator zero, therefore vertical asymptote" without factoring.
The correct way: Factor numerator and denominator first. A factor that cancels leaves a removable hole; a factor that survives gives an infinite discontinuity.
The first instinct on a rational function is to find where the denominator is zero and label it an asymptote — and the cancelling-factor case is exactly where that instinct drops the hole and mislabels the break.
Mistake 2: Assuming "defined at the point" means "continuous there"
Where it slips in: Piecewise functions that assign a value at the switch point, as in Example 3.
Don't do this: See that $h(1) = 4$ exists and conclude the function is continuous at $x = 1$.
The correct way: Check all three conditions. A value can exist and sit off the curve — the limit and the value must also be equal.
The memorizer who learned "discontinuous means undefined" freezes here, because the function is defined at the point. Existence is one of three conditions, not the whole test.
Mistake 3: Mixing up left- and right-hand limits on a jump
Where it slips in: Step functions and piecewise rules at the switch point.
Don't do this: Evaluate only one side and call the limit done.
The correct way: Compute both one-sided limits. A jump is defined by them existing and disagreeing — you cannot detect it from one side.
Practice Questions
Work through these, then check your answers below. For each function, name the point(s) of discontinuity and the type.
Classify the discontinuity of $f(x) = \dfrac{1}{x - 2}$.
Classify the discontinuity of $g(x) = \dfrac{x^2 - 1}{x - 1}$.
Where is the floor function $\lfloor 2x \rfloor$ discontinuous near $x = 1.5$?
Classify the discontinuity of $h(x) = \begin{cases} x + 2, & x < 0 \ x - 2, & x \geq 0 \end{cases}$ at $x = 0$.
Is the polynomial $f(x) = x^3 - 4x$ discontinuous anywhere?
Answers
Answer to Question 1: Infinite discontinuity at $x = 2$. The denominator is zero and does not cancel, so the one-sided limits run to $\pm\infty$.
Answer to Question 2: Removable discontinuity (a hole) at $x = 1$. The factor cancels: $g(x) = \dfrac{(x-1)(x+1)}{x-1} = x + 1$ for $x \neq 1$, leaving a hole at $(1, 2)$.
Answer to Question 3: Jump discontinuity at $x = 1.5$. The left-hand limit is $2$ and the right-hand limit is $3$, since $\lfloor 2x \rfloor$ steps up there.
Answer to Question 4: Jump discontinuity at $x = 0$. The left-hand limit is $2$, the right-hand limit is $-2$, and the two disagree.
Answer to Question 5: No. Every polynomial is continuous on all real numbers, so $f(x) = x^3 - 4x$ has no discontinuities.
Conclusion
A discontinuous function is one whose graph has at least one break — a point where the three-part continuity test fails.
The three types are removable (a hole), jump (a step), and infinite (an asymptote); each has a distinct graph signature.
A zero denominator does not always mean an asymptote — factor first, because a cancelling factor gives a removable hole instead.
"Defined at the point" is only one of three continuity conditions; the limit must exist and equal the value too.
Continuity is the gate for the Intermediate Value Theorem and calculus, so knowing where a function breaks tells you where those tools stop.
A Practical Next Step
Work through the practice questions above to solidify your understanding. If you get stuck on the factoring step, return to Example 2 and re-trace the cancellation before labelling the break. At Bhanzu, trainers teach the three-condition continuity test as a single checklist students run at every candidate point, so the type is never guessed. Want a live Bhanzu trainer to walk through more discontinuous function problems? Book a free demo class.
Was this article helpful?
Your feedback helps us write better content