In short

A function is discontinuous at a point when at least one of the three continuity conditions fails. There are four distinct types: removable (a hole you can fill), jump (the function leaps from one value to another), infinite (the function blows up to \pm\infty), and oscillatory (the function vibrates too wildly to settle on any limit). Each type has a different cause, a different graph, and a different cure — or no cure at all.

You know what continuity means: \lim_{x \to a} f(x) = f(a). Three conditions, all satisfied. But when a function fails this test, the way it fails matters. A tiny repairable hole is a very different problem from a function that shoots off to infinity or one that shakes back and forth without settling.

Consider these four functions, all discontinuous at x = 0:

f_1(x) = \frac{x^2}{x}, \qquad f_2(x) = \text{sgn}(x), \qquad f_3(x) = \frac{1}{x}, \qquad f_4(x) = \sin\!\left(\frac{1}{x}\right)

Each one breaks at the origin, but for a completely different reason. The first has a hole you can patch. The second has a cliff the function jumps across. The third has a vertical explosion. The fourth has an uncontrollable vibration. These are the four types of discontinuity, and by the end of this article you will be able to name each one on sight.

Type 1: Removable discontinuity

Take f(x) = \dfrac{x^2 - 4}{x - 2}. At x = 2, the denominator is zero, so f(2) is undefined. But look at what happens near x = 2. Factor the numerator:

\frac{x^2 - 4}{x - 2} = \frac{(x-2)(x+2)}{x-2} = x + 2 \quad \text{for } x \neq 2

So for every x except 2, the function is just x + 2. A perfectly good straight line. The limit at x = 2 is 4. The graph looks like the line y = x + 2 with a single point drilled out at (2, 4).

The function $f(x) = (x^2 - 4)/(x - 2)$. Everywhere except $x = 2$, it equals $x + 2$. At $x = 2$ there is a hole — the function is not defined there. The limit exists (it is $4$), but there is no function value to match it with.

The fix is obvious: define f(2) = 4. The hole is filled, the function becomes x + 2 everywhere, and continuity is restored. That is why this type is called removable — the discontinuity can be removed by redefining the function at a single point.

Removable discontinuity

f has a removable discontinuity at x = a if \displaystyle\lim_{x \to a} f(x) exists (and is finite), but either f(a) is not defined or f(a) \neq \displaystyle\lim_{x \to a} f(x).

The discontinuity is "removed" by defining (or redefining) f(a) = \displaystyle\lim_{x \to a} f(x).

A removable discontinuity is the mildest kind. The function wants to be continuous — it is heading toward a definite value — but something goes wrong at the single point itself. Either the point is missing from the domain, or someone stuck the wrong value in.

Two common sources:

Type 2: Jump discontinuity

Take the sign function:

\text{sgn}(x) = \begin{cases} -1 & \text{if } x < 0 \\ 0 & \text{if } x = 0 \\ 1 & \text{if } x > 0 \end{cases}

At x = 0, the left-hand limit is -1 and the right-hand limit is +1. The two one-sided limits both exist, but they are not equal. No matter how you define f(0) — whether 0, -1, 1, or anything else — you cannot make both one-sided limits equal to it. The function has a permanent gap that no single-point redefinition can fix.

The sign function $\text{sgn}(x)$ has a jump discontinuity at $x = 0$. The function approaches $-1$ from the left and $+1$ from the right. The value at the origin is $0$, but no choice of value can bridge the gap between the two sides.

Jump discontinuity

f has a jump discontinuity at x = a if both one-sided limits exist and are finite, but they are not equal:

\lim_{x \to a^-} f(x) \neq \lim_{x \to a^+} f(x)

The size of the jump is \left|\lim_{x \to a^+} f(x) - \lim_{x \to a^-} f(x)\right|.

Jump discontinuities are the most visually obvious type. The graph has a cliff: the function is at one height on the left, a different height on the right, and no smooth transition between them.

The floor function \lfloor x \rfloor has a jump discontinuity at every integer. At x = 3, for instance, the left-hand limit is 2 and the right-hand limit is 3. The size of the jump is 1.

Jump discontinuities cannot be repaired by redefining the function at a single point. You can change f(a) to be the left limit, the right limit, their average, or anything you like — but the two one-sided limits will still disagree with each other, so the two-sided limit will still not exist. The discontinuity is essential, not removable.

Here is another example that appears constantly in JEE-level problems. Consider the function:

f(x) = \begin{cases} x + 1 & \text{if } x < 3 \\ x^2 - 5 & \text{if } x \geq 3 \end{cases}

From the left: \lim_{x \to 3^-} (x + 1) = 4. From the right: \lim_{x \to 3^+} (x^2 - 5) = 4. Both limits equal 4, and f(3) = 9 - 5 = 4. So f is continuous at x = 3 — the pieces join perfectly.

Now change the second piece to x^2 - 4:

g(x) = \begin{cases} x + 1 & \text{if } x < 3 \\ x^2 - 4 & \text{if } x \geq 3 \end{cases}

From the left: \lim_{x \to 3^-} (x + 1) = 4. From the right: \lim_{x \to 3^+} (x^2 - 4) = 5. The one-sided limits are 4 and 5 — different. The function jumps from height 4 to height 5 at x = 3. This is a jump discontinuity with jump size |5 - 4| = 1.

The message: whether a piecewise function has a jump discontinuity at the boundary depends entirely on whether the two pieces arrive at the same height. If they do, continuous. If they do not, jump.

Type 3: Infinite discontinuity

Take f(x) = \dfrac{1}{x} at x = 0.

As x approaches 0 from the right, f(x) = 1/x grows without bound: 1/0.1 = 10, 1/0.01 = 100, 1/0.001 = 1000, and so on toward +\infty. From the left, it plunges toward -\infty: 1/(-0.1) = -10, 1/(-0.01) = -100, and so on.

Neither one-sided limit is a finite number. The limit does not exist — not because it fails to settle (as in the oscillatory case below), but because it escapes to infinity.

The function $f(x) = 1/x$ has an infinite discontinuity at $x = 0$. As $x$ approaches zero from the right, $f(x)$ shoots to $+\infty$; from the left, to $-\infty$. The vertical axis is a vertical asymptote.

Infinite discontinuity

f has an infinite discontinuity at x = a if at least one of the one-sided limits is +\infty or -\infty:

\lim_{x \to a^+} f(x) = \pm\infty \quad \text{or} \quad \lim_{x \to a^-} f(x) = \pm\infty

The vertical line x = a is called a vertical asymptote of f.

Infinite discontinuities typically occur where the function has a denominator that becomes zero while the numerator stays nonzero. Some further examples:

An infinite discontinuity is never removable. You cannot assign a real-number value to f(a) that makes the limit equal the value, because the limit is not a real number. The function is fundamentally blowing up.

Type 4: Oscillatory discontinuity

This is the strangest of the four types. Take f(x) = \sin(1/x) at x = 0.

As x gets small, 1/x gets large, and \sin(1/x) oscillates between -1 and +1 faster and faster. At x = 1/\pi, the argument is \pi, and \sin \pi = 0. At x = 1/(2\pi), the argument is 2\pi, and \sin 2\pi = 0 again. But between those two values, the sine has completed a full cycle. As x shrinks further, the cycles get compressed into tinier and tinier intervals. The oscillations never stop, never slow down, and never settle toward any single value.

The function $\sin(1/x)$ near $x = 0$. As $x$ approaches zero, the oscillations become infinitely rapid. The function visits every value between $-1$ and $+1$ infinitely many times in every neighbourhood of zero — there is no single value it approaches.

The limit \lim_{x \to 0} \sin(1/x) does not exist. Not because the function heads to infinity (it stays between -1 and +1), but because it cannot commit to any particular value. It keeps changing its mind, oscillating without settling.

Oscillatory discontinuity

f has an oscillatory discontinuity at x = a if f is bounded near a but \displaystyle\lim_{x \to a} f(x) does not exist because f oscillates between different values without converging.

Oscillatory discontinuities are rarer than the other three types in school-level problems. They require functions with an argument that changes infinitely fast — like \sin(1/x) or \cos(1/x) at the origin. But they are conceptually important because they show that "the limit does not exist" can mean something subtler than "the function blows up."

There is a related function that is worth seeing: f(x) = x \sin(1/x). This function also oscillates infinitely rapidly near x = 0, but the oscillations are dampened by the factor of x. As x \to 0, the amplitude of the oscillation shrinks to zero, and \lim_{x \to 0} x \sin(1/x) = 0. If you define f(0) = 0, this function is actually continuous at x = 0. The dampening tames the oscillation.

The function $x\sin(1/x)$ (black) is squeezed between $y = x$ and $y = -x$ (dashed). As $x \to 0$, the envelope closes down to zero, forcing the limit to be $0$. This is the sandwich theorem in action.

Putting it all together

Here is a summary. At any point where f is discontinuous, exactly one of the following is true:

Type What happens Limit exists? Fixable?
Removable Hole or wrong value; limit exists and is finite Yes Yes — redefine f(a)
Jump Left and right limits are different finite numbers No (one-sided limits exist, two-sided does not) No
Infinite At least one side goes to \pm\infty No No
Oscillatory Function oscillates without settling No No

The first type is a first kind discontinuity where the repair is trivial. The second type is also called a first kind discontinuity (both one-sided limits exist). The third and fourth types are called second kind discontinuities — at least one of the one-sided limits fails to exist as a finite value.

Worked examples

Example 1: Classify the discontinuity of f(x) = (sin x)/x at x = 0

Step 1. Check whether f(0) is defined.

f(0) = \sin(0)/0 = 0/0, which is undefined. So condition 1 of continuity fails.

Why: division by zero means the function has no value at this point.

Step 2. Compute the limit.

\lim_{x \to 0} \frac{\sin x}{x} = 1

This is a standard limit. It can be proved using the sandwich theorem with the inequalities \cos x \leq \frac{\sin x}{x} \leq 1 for small positive x.

Why: the limit exists and equals a specific finite number, even though the function is undefined at x = 0.

Step 3. Compare with the definition of each type.

The limit exists and is finite (=1), but f(0) is not defined. This matches the definition of a removable discontinuity.

Why: the function is "almost continuous" — it only fails because of a single missing point.

Step 4. State the repair.

Define f(0) = 1, and the function becomes continuous at x = 0.

Why: after this redefinition, \lim_{x \to 0} f(x) = 1 = f(0), so all three conditions hold.

Result: Removable discontinuity at x = 0. The repaired function, f(x) = (\sin x)/x for x \neq 0 and f(0) = 1, is called the sinc function and is continuous everywhere.

The function $(\sin x)/x$. The curve approaches $1$ as $x \to 0$ from both sides. The red dot at $(0, 1)$ shows where the function should be defined to fill the hole and make it continuous.

The graph is a smooth wave that decays as you move away from the origin. At x = 0, the hole is invisible to the eye — the graph looks like it passes through (0, 1) without any break. That is exactly the point: a removable discontinuity is so mild that you cannot see it on a graph.

Example 2: Classify the discontinuity of a piecewise function

Consider:

f(x) = \begin{cases} x + 1 & \text{if } x < 1 \\ 3 & \text{if } x = 1 \\ x^2 + 1 & \text{if } x > 1 \end{cases}

Is f continuous at x = 1? If not, what type of discontinuity?

Step 1. Check that f(1) is defined.

f(1) = 3 (from the middle piece).

Why: the function explicitly assigns a value at x = 1.

Step 2. Compute the left-hand limit.

\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x + 1) = 2

Why: for x < 1, the function is x + 1. As x approaches 1 from the left, this approaches 2.

Step 3. Compute the right-hand limit.

\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x^2 + 1) = 1 + 1 = 2

Why: for x > 1, the function is x^2 + 1. As x approaches 1 from the right, this also approaches 2.

Step 4. Both one-sided limits equal 2, so \lim_{x \to 1} f(x) = 2. Compare with f(1) = 3.

\lim_{x \to 1} f(x) = 2 \neq 3 = f(1)

Why: the limit exists and is finite, but it disagrees with the function's value. This is a removable discontinuity — the function is defined at the wrong height.

Result: Removable discontinuity at x = 1. Redefine f(1) = 2 to repair it.

The piecewise function near $x = 1$. Both pieces approach the height $2$ as $x \to 1$, but the assigned value $f(1) = 3$ sits one unit higher. The filled red dot is the function's actual value; the open circle at $(1, 2)$ marks where the graph "wants" to be. Move the red dot down to $(1, 2)$ and the discontinuity vanishes.

The two pieces of the function — the line x + 1 and the parabola x^2 + 1 — both arrive at height 2 as x approaches 1. The discontinuity exists only because f(1) was assigned the value 3 instead of 2. This is the hallmark of a removable discontinuity: the surrounding behaviour is perfectly well-behaved, and only the single point is wrong.

Common confusions

Going deeper

If you can identify the four types of discontinuity and check which type applies to a given function, you have everything this article set out to teach. The rest is for readers who want the formal classification and a subtlety about "first kind" versus "second kind."

First kind and second kind

Indian textbooks (and many international ones) use a binary classification alongside the four-type one:

Discontinuity of the first kind at x = a: both \lim_{x \to a^-} f(x) and \lim_{x \to a^+} f(x) exist as finite numbers, but f is still discontinuous at a. This covers two of the four types:

A first-kind discontinuity where the two one-sided limits are equal is sometimes called a removable discontinuity of the first kind. A first-kind discontinuity where they differ is called a non-removable discontinuity of the first kind or simply a jump.

Discontinuity of the second kind at x = a: at least one of the one-sided limits does not exist as a finite number. This covers:

The terminology "first kind" and "second kind" is especially common in NCERT and ISC textbooks. You should know both systems.

Can a function have uncountably many discontinuities?

The functions you meet in school typically have finitely many discontinuities (like a piecewise function) or countably many (like the floor function, which has one at every integer). But it is possible to construct a function with uncountably many discontinuities — the Dirichlet function D(x), which equals 1 if x is rational and 0 if x is irrational, is discontinuous at every real number. It has as many discontinuities as there are points on the real line.

Conversely, a theorem of Froda says that a monotonic function (one that is always increasing or always decreasing) can have at most countably many discontinuities, and all of them must be of the jump type. So the structure of a function constrains what kinds of discontinuities it can have.

These facts are beyond school syllabus, but they give you a sense of why the classification matters: it tells you something deep about the function's structure, not just a surface-level description of its graph.

A decision tree for classifying discontinuities

When you encounter a function that is discontinuous at x = a, here is the systematic way to classify it:

Step 1. Compute \lim_{x \to a^-} f(x) and \lim_{x \to a^+} f(x) separately.

Step 2. If either one-sided limit does not exist as a finite number:

Step 3. If both one-sided limits exist and are finite:

This flowchart handles every discontinuity you will meet in school and competitive exams. Memorize the tree: one-sided limits first, then compare.

A subtle case: removable discontinuity versus no discontinuity

It is worth being precise about the boundary between "removable discontinuity" and "continuous." Consider:

f(x) = \begin{cases} \dfrac{x^2 - 1}{x - 1} & \text{if } x \neq 1 \\[6pt] 2 & \text{if } x = 1 \end{cases}

For x \neq 1: \dfrac{x^2 - 1}{x - 1} = \dfrac{(x-1)(x+1)}{x-1} = x + 1. The limit at x = 1 is 2. And f(1) = 2. So \lim_{x \to 1} f(x) = 2 = f(1) — the function is continuous at x = 1. No discontinuity at all, despite the suspicious-looking formula.

Now change the value: f(1) = 7. The limit is still 2, but f(1) = 7 \neq 2 — removable discontinuity.

And without any definition at all — just f(x) = (x^2 - 1)/(x - 1) — the function is undefined at x = 1. The limit exists (= 2), but there is no value to compare it with — also a removable discontinuity, though for a different reason (condition 1 fails instead of condition 3).

The distinction matters when JEE problems ask "for what value of k is f continuous at x = a?" — they are asking you to find the value that removes the removable discontinuity.

Where this leads next