Removable Discontinuity: Definition & How To Find It

What Is A Removable Discontinuity?

A removable discontinuity is a point where a function's graph has a single hole: the curve heads toward one clear height from both sides, but the point itself is missing or misplaced. It is the gentlest way a function can fail to be continuous, because everything around the point behaves, and only the one point is wrong.

A function $f$ has a removable discontinuity at $x = a$ exactly when the two conditions below both hold.

$$\begin{aligned} &\text{(1)} \quad \lim_{x \to a} f(x) = L \ \text{exists and is finite} \ &\text{(2)} \quad f(a) \ \text{is undefined, or } f(a) \neq L \end{aligned}$$

Condition (1) is what makes the gap fixable: the limit of a function already agrees on a single value $L$ from the left and the right. Condition (2) is what makes it a discontinuity in the first place: the plotted value is missing or sits at the wrong height. When both hold, the graph shows an open circle at $(a, L)$, and sometimes a stray solid dot elsewhere.

This is one of the named types of discontinuity. It is the only removable one. The others, where the limit itself fails to exist, cannot be repaired by touching a single point.

How Do You Find A Removable Discontinuity?

For a rational function, a removable discontinuity appears wherever a factor cancels between the numerator and the denominator. The recipe is short.

• Factor the numerator and the denominator completely.

• Cancel any factor common to both. The input that makes a cancelled factor zero is the location $x = a$ of the hole.

• Substitute that $x$-value into the simplified expression. The result is $L$, the height of the hole at $(a, L)$.

The three worked examples below are the cases you meet most: a linear hole, a second linear hole, and the famous trigonometric one.

Example 1: Find the removable discontinuity of $f(x) = \dfrac{x^2 - 1}{x - 1}$.

Substituting $x = 1$ gives $\dfrac{0}{0}$, so $f(1)$ is undefined. Factor and cancel for $x \neq 1$:

$$\frac{x^2 - 1}{x - 1} = \frac{(x - 1)(x + 1)}{x - 1} = x + 1$$

The cancelled factor $(x - 1)$ is zero at $x = 1$, so the hole sits there. Substitute into the simplified form:

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

The limit is $2$ but $f(1)$ does not exist, so $f$ has a removable discontinuity at $x = 1$.

Final answer: Hole at $(1, 2)$; the limit is $2$ and $f(1)$ is undefined.

Example 2: Find the removable discontinuity of $f(x) = \dfrac{x^2 - 4}{x - 2}$.

Again substitution gives $\dfrac{0}{0}$ at $x = 2$. Factor and cancel:

$$\frac{x^2 - 4}{x - 2} = \frac{(x - 2)(x + 2)}{x - 2} = x + 2, \qquad \lim_{x \to 2} f(x) = 2 + 2 = 4$$

Final answer: Hole at $(2, 4)$; the limit is $4$ and $f(2)$ is undefined.

Example 3: Show $f(x) = \dfrac{\sin x}{x}$ has a removable discontinuity at $x = 0$.

Here $f(0) = \dfrac{0}{0}$ is undefined, and no algebraic factor cancels. Instead this relies on the standard limit

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

a known result that the squeeze theorem establishes. Because the limit exists ($1$) while $f(0)$ is missing, the point is a removable discontinuity.

Final answer: Hole at $(0, 1)$; defining $f(0) = 1$ produces the continuous sinc function.

How Do You Remove A Removable Discontinuity?

You remove the discontinuity by redefining the function at the single bad point so that its value equals the limit. Formally, replace $f$ with the patched function

$$\tilde{f}(x) = \begin{cases} f(x), & x \neq a \ L, & x = a \end{cases} \qquad \text{where } L = \lim_{x \to a} f(x).$$

Now all three conditions for continuity of a function hold at $a$: the value exists, the limit exists, and the two agree. The rest of the graph never changes, because $\tilde{f}$ and $f$ differ at exactly one input.

For Example 1, the patched function is a piecewise function:

$$\tilde{f}(x) = \begin{cases} \dfrac{x^2 - 1}{x - 1}, & x \neq 1 \[4pt] 2, & x = 1 \end{cases}$$

which is simply the line $y = x + 1$ with no hole left. This one-point repair is the whole reason the word "removable" is used, and it is the exact skill AP Calculus lists under "removing a discontinuity."

What Is The Difference Between A Removable And A Non-Removable Discontinuity?

The dividing line is one question: does the two-sided limit exist? If it does, the discontinuity is removable. If it does not, no single value can repair it, and the discontinuity is non-removable.

Non-removable discontinuities come in two standard kinds. A jump discontinuity has left and right limits that both exist but disagree, so the graph steps from one height to another. An infinite discontinuity has a limit that runs off to $\pm\infty$, producing a vertical asymptote.

Table: How the three discontinuity types compare, and which can be repaired.

The contrast with the vertical asymptote is the one that trips students, because both a hole and an asymptote come from a zero denominator. The test is whether the offending factor cancels. A factor of the denominator that cancels with the numerator leaves a hole; a factor that does not cancel forces an asymptote. For $\dfrac{1}{x - 2}$ nothing cancels, the limit at $2$ is infinite, and the break is non-removable.

Why Is It Called A "Removable" Discontinuity?

The name is literal, not decorative, and the intuition is worth stating plainly.

• The limit already carries the answer. When $\lim_{x \to a} f(x) = L$ exists, the function is telling you the exact height it wants at $a$. The only thing missing is the instruction to plot it there.

• Only one point is wrong. Every other input behaves. Changing $f$ at a single value cannot disturb the neighbours, because a limit depends on what happens near $a$, never on the value at $a$ itself.

• Non-removable breaks fail this test. A jump or an asymptote has no single "right height" to drop in, since the two sides disagree or run to infinity. There is nothing to fill, which is why those stay broken.

Seen this way, removability is a statement about information: the graph has already computed the value it needs, so the discontinuity is a clerical gap rather than a genuine tear in the curve.

Who Discovered The Idea Of A Removable Discontinuity?

Classifying breaks in a graph only became possible once "the limit exists but the value is missing" could be stated precisely, which took the arrival of a rigorous limit in the 1800s.

Two names anchor the rigor behind the definition:

• Augustin-Louis Cauchy (1789–1857, France) gave continuity and the limit their careful definitions in his Cours d'analyse of 1821, making "the limit exists but $f(a)$ does not" a statement you could prove rather than merely sketch.

• Karl Weierstrass (1815–1897, Germany) sharpened those definitions into the precise epsilon-delta language that lets us separate a removable hole from a genuine singularity without appeal to a picture.

Where Are Removable Discontinuities Used In The Real World?

A removable discontinuity is exactly the situation where a formula breaks at one input but the quantity it models is perfectly well-defined there, so filling the hole is routine across the sciences.

• Signal processing: the sinc function $\dfrac{\sin x}{x}$, patched to equal $1$ at $0$, is the ideal interpolation kernel used to reconstruct a continuous signal from its samples. The single filled hole makes the whole reconstruction smooth.

• Physics and engineering: many formulas take a $\dfrac{0}{0}$ form at a special value (a rate at zero speed, a field on an axis of symmetry) where the true physical quantity is finite. Engineers "remove" the hole by using the limiting value.

• Computer graphics and CAD: when two surface patches are stitched, an isolated undefined point at a seam is filled with its limit so the rendered surface shows no puncture.

• Numerical computing: software that would divide by zero at one input substitutes the limit instead, avoiding a crash while keeping the output exact everywhere the formula is valid.

Across all of these the move is identical to the classroom one: the surrounding values already fix the answer, so the isolated gap is filled rather than feared.

What Are The Most Common Mistakes With Removable Discontinuity?

These three errors account for most lost marks on this topic, and each matches a question real students ask on r/calculus, Quora, and AP review guides such as Albert.io's discontinuity walkthroughs.

Confusing a hole with a vertical asymptote.

Where it slips in:

A student sees a zero denominator and assumes an asymptote, or sees a hole and assumes the graph is smooth there, without factoring first.

Don't do this:

Do not classify from the denominator alone. A zero denominator can give either a removable hole or a non-removable asymptote.

The correct way:

Factor the numerator and denominator completely first. A factor that cancels leaves a hole; a factor that does not cancel forces a vertical asymptote. For $\dfrac{x^2 - 1}{x - 1}$ the $(x-1)$ cancels, so $x = 1$ is a hole, not an asymptote.

Thinking the function is continuous as it stands.

Where it slips in:

A student computes the limit, finds it exists, and concludes the function is already continuous at that point.

Don't do this:

Do not treat a healthy limit as continuity. A removable discontinuity has a perfectly good limit and is still discontinuous until the value is defined or corrected.

The correct way:

Compare the limit with the value. Continuity needs $f(a) = \lim_{x \to a} f(x)$; if $f(a)$ is missing or unequal, the function is discontinuous until you patch it with $\tilde{f}(a) = L$.

Plotting the point at $f(a)$ instead of at the limit.

Where it slips in:

A student marks the hole at whatever the original expression seems to give, or at a stray defined value, rather than at the height the curve approaches.

Don't do this:

Do not read the hole's height off the unsimplified formula or off a misplaced dot. The hole sits at the limit, not at the broken value.

The correct way:

Substitute into the simplified expression to get $L$, then place the open circle at $(a, L)$. For $\dfrac{x^2 - 4}{x - 2}$ the hole is at $(2, 4)$, the value of $x + 2$ at $x = 2$.

Practice Problems On Removable Discontinuity

Work each one, then check against the answer. Answers are verified.

1. Find the removable discontinuity of $f(x) = \dfrac{x^2 - 9}{x - 3}$.
   (Answer: cancels to $x + 3$; hole at $(3, 6)$.)

2. Does $f(x) = \dfrac{x - 5}{x^2 - 25}$ have a removable discontinuity, and where?
   (Answer: yes; $\dfrac{x-5}{(x-5)(x+5)} = \dfrac{1}{x+5}$, hole at $\left(5, \tfrac{1}{10}\right)$. There is also a non-removable asymptote at $x = -5$.)

3. Redefine $f(x) = \dfrac{x^2 - x - 6}{x - 3}$ to make it continuous at $x = 3$.
   (Answer: cancels to $x + 2$, limit $5$, so set $f(3) = 5$.)

4. Is the break in $f(x) = \dfrac{1}{x - 4}$ removable?
   (Answer: no; the factor does not cancel, the limit is infinite, so it is a non-removable infinite discontinuity.)

5. Find the hole of $f(x) = \dfrac{x^3 - 1}{x - 1}$.
   (Answer: cancels to $x^2 + x + 1$; hole at $(1, 3)$.)

6. For $f(x) = \dfrac{\sin 2x}{x}$, find the value that removes the discontinuity at $x = 0$.
   (Answer: $\lim_{x \to 0} \dfrac{\sin 2x}{x} = 2$, so define $f(0) = 2$.)

Where Should You Go Next After Removable Discontinuity?

Removable discontinuity is one branch of a larger map, and several natural doors open from here.

1. Types of discontinuity. See all three kinds, removable, jump, and infinite, sorted side by side with the test for each.

2. Jump discontinuity. Study the non-removable break where the left and right limits disagree.

3. Continuity and differentiability. See why continuity is the entry requirement for a derivative, and what a patched hole buys you.

If your child is meeting discontinuities for the first time, a live Bhanzu trainer teaches the hole-versus-asymptote test from the limit up, so limits, continuity, and derivatives feel like one connected story, in the Bhanzu math program.