What Is Infinite Discontinuity?
An infinite discontinuity is a point where a function fails to be continuous because at least one of its one-sided limits is infinite. Written with limits, a function $f$ has an infinite discontinuity at $x = a$ when
$$\lim_{x \to a^{-}} f(x) = \pm\infty \quad \text{or} \quad \lim_{x \to a^{+}} f(x) = \pm\infty.$$
The symbol $\pm\infty$ is not a number the graph reaches. It is shorthand for "grows without bound," so the two-sided limit $\lim_{x \to a} f(x)$ does not exist as a finite value. Geometrically, the curve hugs a vertical asymptote: a vertical line $x = a$ that the graph runs alongside, climbing toward the top of the plane or plunging toward the bottom, without ever crossing it.
This is one of the three standard kinds of break a function can have. For the full map of all of them, see types of discontinuity, and for the property this one destroys, see continuity of a function. What marks an infinite discontinuity out is that the trouble is unbounded size, not a mismatch or a hole.
How Do You Identify An Infinite Discontinuity?
The fastest test lives in the denominator. For a rational function, an infinite discontinuity appears where the denominator hits zero but the numerator does not, after you have cancelled every common factor.
Denominator zero, numerator non-zero: the fraction blows up, and $x = a$ is an infinite discontinuity with a vertical asymptote.
Denominator zero and numerator zero: a factor may cancel, leaving a finite limit and a fixable hole instead. That is a removable discontinuity, not an infinite one.
The cancel-first step is the whole game. Reading a zero in the denominator as an automatic asymptote is the most common error on this topic, and it has its own entry in the mistakes section below. For the cancelling itself, the algebra of simplifying rational expressions is the tool you use.
There are two visual shapes an infinite discontinuity can take, and telling them apart is worth a moment:
Both one-sided limits run the same way. For $\frac{1}{(x-a)^2}$, both sides climb to $+\infty$, so the graph makes a matching pair of upward spikes against the asymptote.
The two sides run opposite ways. For $\frac{1}{x}$, the left side dives to $-\infty$ while the right side climbs to $+\infty$, so the graph splits across the asymptote.
Either shape counts as an infinite discontinuity, because the defining condition asks only that at least one side be infinite.
What Are Examples Of Infinite Discontinuity?
Four functions cover almost every infinite discontinuity you will meet. Each one-sided limit below is exact.
The reciprocal $\frac{1}{x}$ at $x = 0$. The two sides disagree in sign:
$$\lim_{x \to 0^{-}} \frac{1}{x} = -\infty, \qquad \lim_{x \to 0^{+}} \frac{1}{x} = +\infty.$$
The even reciprocal $\frac{1}{(x-a)^2}$ at $x = a$. The square keeps the denominator positive on both sides, so both limits agree:
$$\lim_{x \to a^{-}} \frac{1}{(x-a)^2} = +\infty, \qquad \lim_{x \to a^{+}} \frac{1}{(x-a)^2} = +\infty.$$
The tangent $\tan x$ at $x = \frac{\pi}{2}$. Since $\tan x = \frac{\sin x}{\cos x}$ and $\cos x \to 0$ there while $\sin x \to 1$:
$$\lim_{x \to (\pi/2)^{-}} \tan x = +\infty, \qquad \lim_{x \to (\pi/2)^{+}} \tan x = -\infty.$$
The tangent repeats this at every $x = \frac{\pi}{2} + n\pi$, one infinite discontinuity per period.
The logarithm $\ln x$ at $x = 0$. Here the function is only defined for $x > 0$, so a single one-sided limit does the work:
$$\lim_{x \to 0^{+}} \ln x = -\infty.$$
One infinite side is enough for the definition, so $\ln x$ has an infinite discontinuity at the edge of its domain. The same idea drives every logarithm; see logarithmic functions for the wider behaviour.
How Does Infinite Discontinuity Differ From Removable And Jump Discontinuity?
All three are points where a function is not continuous, but the reason differs, and that reason is what you name.
Table: The three standard discontinuities, told apart by their one-sided limits.
Type | One-sided limits | Two-sided limit | Can it be patched? |
|---|---|---|---|
Both finite and equal | Exists (finite) | Yes, redefine one point | |
Both finite but unequal | Does not exist | No | |
Infinite | At least one is $\pm\infty$ | Does not exist | No |
The line that trips students is the one between jump and infinite. In both, the two-sided limit fails to exist, so "the limit does not exist" alone does not tell them apart. The extra question is why it fails: a jump fails because two finite heights disagree, while an infinite discontinuity fails because the height itself is unbounded.
Only the infinite case produces a vertical asymptote. For the machinery underneath all of this, the limit of a function is the place to start.
How Do You Work With Infinite Discontinuities? Worked Examples
Each example checks the one-sided limits directly, so nothing rests on a memorised rule.
Example 1: A clean rational blow-up.
Classify the discontinuity of $f(x) = \dfrac{1}{x - 3}$ at $x = 3$.
The denominator is zero at $x = 3$, and the numerator is $1 \neq 0$, so no factor cancels. Check each side:
$$\lim_{x \to 3^{-}} \frac{1}{x-3} = -\infty, \qquad \lim_{x \to 3^{+}} \frac{1}{x-3} = +\infty.$$
Both sides are infinite, so this is an infinite discontinuity with a vertical asymptote at $x = 3$.
Final answer: infinite discontinuity at $x = 3$; vertical asymptote $x = 3$.
Example 2: Removable and infinite in the same function.
Classify every discontinuity of $f(x) = \dfrac{x + 2}{x^2 - 4}$.
Factor first: $x^2 - 4 = (x-2)(x+2)$, so
$$f(x) = \frac{x+2}{(x-2)(x+2)} = \frac{1}{x-2}, \quad x \neq -2.$$
At $x = -2$ the common factor $(x+2)$ cancelled, so the limit is finite:
$$\lim_{x \to -2} \frac{1}{x-2} = \frac{1}{-4} = -0.2500.$$
That is a removable discontinuity, a fixable hole, not an infinite one. At $x = 2$ nothing cancels, the numerator $1$ is non-zero, and
$$\lim_{x \to 2^{-}} \frac{1}{x-2} = -\infty, \qquad \lim_{x \to 2^{+}} \frac{1}{x-2} = +\infty.$$
Final answer: removable discontinuity at $x = -2$; infinite discontinuity (vertical asymptote) at $x = 2$.
Example 3: A matching-spike blow-up.
Classify the discontinuity of $f(x) = \dfrac{5}{(x-1)^2}$ at $x = 1$.
The denominator $(x-1)^2$ is zero at $x = 1$ and positive on both sides, while the numerator is $5 \neq 0$. So both one-sided limits climb the same way:
$$\lim_{x \to 1^{-}} \frac{5}{(x-1)^2} = +\infty, \qquad \lim_{x \to 1^{+}} \frac{5}{(x-1)^2} = +\infty.$$
Final answer: infinite discontinuity at $x = 1$; the graph spikes to $+\infty$ on both sides of the asymptote $x = 1$.
What Does A Table Of Common Infinite Discontinuities Look Like?
Table: Standard functions with an infinite discontinuity, and the one-sided limits that prove it.
Function | Discontinuity at | Left limit | Right limit | Shape |
|---|---|---|---|---|
$\dfrac{1}{x}$ | $x = 0$ | $-\infty$ | $+\infty$ | Split |
$\dfrac{1}{(x-a)^2}$ | $x = a$ | $+\infty$ | $+\infty$ | Matching spikes |
$\tan x$ | $x = \frac{\pi}{2} + n\pi$ | $+\infty$ | $-\infty$ | Split |
$\ln x$ | $x = 0$ | undefined | $-\infty$ | One-sided edge |
$\dfrac{1}{x-3}$ | $x = 3$ | $-\infty$ | $+\infty$ | Split |
Read the table by the limits, not the picture. Whenever a column shows $\pm\infty$, the point is an infinite discontinuity and a vertical asymptote sits there. These asymptotic tails are also why the behaviour far out matters; the companion idea of a horizontal tail is covered in limits at infinity.
Why Does Infinite Discontinuity Happen?
An infinite discontinuity is what a function does when it is asked to divide a fixed amount by something shrinking to nothing.
Small divisors make big results. In $\frac{1}{x}$, as $x$ shrinks toward $0$, you are splitting $1$ into ever tinier pieces, so the count of pieces races upward. The closer $x$ gets to the forbidden value, the more extreme the output, and there is no largest value it settles on.
The sign is decided by the approach. With $\frac{1}{x}$, a small negative $x$ gives a large negative result and a small positive $x$ gives a large positive one, which is why the two sides split. With $\frac{1}{(x-a)^2}$ the square makes every nearby divisor positive, so both sides agree and spike together.
The asymptote is the boundary of the domain. The function is simply not defined at $x = a$, and the vertical line $x = a$ marks that missing input. The graph can approach the line as closely as you like but never lands on it, which is the visual signature of an infinite limit.
Seen this way, an infinite discontinuity is not an exotic defect. It is the ordinary result of division meeting a zero it cannot swallow. These functions are usually rational functions or ratios like $\tan x$, where a denominator is genuinely allowed to reach zero.
Who Shaped The Rigorous Definition Of Discontinuity?
For a long time "continuous" meant only "drawable without lifting the pen," a picture with no precise test behind it. The modern definition, and with it a clean way to name an infinite discontinuity, came from the nineteenth-century drive to put calculus on solid ground.
Two named figures anchor the story:
Bernard Bolzano (1781–1848, Bohemia, then part of the Austrian Empire) gave an early limit-based definition of continuity that went largely unnoticed in his lifetime.
Karl Weierstrass (1815–1897, Germany) later supplied the fully formal $\varepsilon$–$\delta$ machinery, closing the last gaps in Cauchy's account and giving the definition the exact form used in classrooms now.
Where Is Infinite Discontinuity Used In The Real World?
Blow-ups to infinity are not just classroom curiosities. They mark the places where a physical model reaches a limit it cannot cross.
Gravity and electric fields: the inverse-square laws for gravitational and electric force behave like $\frac{1}{r^2}$, so the modelled force has an infinite discontinuity as the distance $r$ to a point mass or point charge shrinks to zero. The infinity is a signal that the "point" idealisation has broken down.
Resonance in engineering: the amplitude of an ideal, undamped oscillator driven at its natural frequency grows without bound, an infinite discontinuity in the response curve that warns designers to add damping to bridges, buildings, and circuits.
Optics and signal strength: brightness and signal intensity fall off like $\frac{1}{r^2}$ from a source, so the same reciprocal-square blow-up appears whenever a model pushes the distance to zero.
Computing and graphics: a division whose denominator can reach zero is a latent infinite discontinuity, which is why renderers, physics engines, and spreadsheets guard against dividing by a vanishing quantity before it returns an infinity or an error.
Across these fields the message is the same: an infinite value in the mathematics is nature or the software telling you a boundary of the model has been hit.
What Are The Most Common Mistakes With Infinite Discontinuity?
These three errors account for most lost marks, and each matches a question real students ask on r/calculus, Google's "People also ask," and course handouts such as Purdue's MA161 notes.
Calling every zero denominator an infinite discontinuity.
Where it slips in:
A student sees $x^2 - 4$ in the denominator, spots that it is zero at $x = 2$ and $x = -2$, and declares a vertical asymptote at both.
Don't do this:
Do not judge from the denominator alone. A zero denominator with a zero numerator may cancel to a finite limit.
The correct way:
Factor and cancel first. For $\frac{x+2}{x^2-4}$, the point $x = -2$ cancels to a removable hole with limit $-0.2500$, and only $x = 2$ is an infinite discontinuity.
Confusing an infinite discontinuity with a jump.
Where it slips in:
A student checks the two-sided limit, sees it does not exist, and writes "jump discontinuity" without looking closer.
Don't do this:
Do not stop at "the limit does not exist." A jump and an infinite discontinuity both fail that test for different reasons.
The correct way:
Read the one-sided limits. Two finite but unequal values is a jump; at least one value of $\pm\infty$ is an infinite discontinuity with a vertical asymptote.
Integrating straight across the asymptote.
Where it slips in:
A student evaluates $\displaystyle\int_{-1}^{1}\frac{1}{x^2},dx$ as $\left[-\frac{1}{x}\right]_{-1}^{1} = -1 - 1 = -2$.
Don't do this:
Do not apply the Fundamental Theorem of Calculus when the integrand has an infinite discontinuity inside the interval. Here $\frac{1}{x^2}$ blows up at $x = 0$.
The correct way:
Notice the infinite discontinuity first. The integrand is positive everywhere, so a negative answer is impossible; the integral is improper and in fact diverges, so no finite value is valid.
Practice Problems On Infinite Discontinuity
Work each one, then check against the answer. Every answer is verified.
Where does $f(x) = \dfrac{1}{x+5}$ have an infinite discontinuity?
(Answer: $x = -5$; a vertical asymptote, with the left limit $-\infty$ and the right limit $+\infty$.)Classify each discontinuity of $f(x) = \dfrac{x-3}{x^2-9}$.
(Answer: factor to $\frac{1}{x+3}$; removable hole at $x = 3$ with limit $\frac{1}{6}$, infinite discontinuity at $x = -3$.)List all infinite discontinuities of $\tan x$.
(Answer: $x = \frac{\pi}{2} + n\pi$ for every integer $n$.)Classify the discontinuity of $f(x) = \dfrac{5}{(x-1)^2}$ at $x = 1$.
(Answer: infinite discontinuity; both one-sided limits are $+\infty$.)Does $f(x) = \dfrac{x^2 - 1}{x - 1}$ have an infinite discontinuity at $x = 1$?
(Answer: no; it cancels to $x + 1$, giving a removable hole with limit $2$.)What kind of discontinuity does $f(x) = \ln(x - 2)$ have at $x = 2$?
(Answer: infinite discontinuity, since $\lim_{x \to 2^{+}} \ln(x-2) = -\infty$; vertical asymptote $x = 2$.)
Where Should You Go Next After Infinite Discontinuity?
Infinite discontinuity sits inside the wider study of limits and continuity, and several natural doors open from here.
Types of discontinuity. See all three breaks side by side, so infinite, removable, and jump each have a clear place.
Limits at infinity. The companion idea: instead of the output running to infinity, the input does, which gives horizontal rather than vertical asymptotes.
Continuity of a function. Go back to the property an infinite discontinuity breaks, and the three-part test that defines it.
If your child is meeting infinite discontinuity for the first time, a live Bhanzu trainer teaches it from the one-sided-limit test up, so the vertical asymptote feels inevitable rather than mysterious, in the Bhanzu math program.
Was this article helpful?
Your feedback helps us write better content