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Removable Discontinuity: A 'Hole' Where the Limit Exists

Removable Discontinuity
Hole

A hole in a graph. That is, a discontinuity that can be "repaired" by filling in a single point. In other words, a removable discontinuity is a point at which a graph is not connected but can be made connected by filling in a single point.

Formally, a removable discontinuity is one at which the limit of the function exists but does not equal the value of the function at that point; this may be because the function does not exist at that point.

Smooth curve with an open circle at (1, 1.25) where the limit lives, and a separate filled dot at (1, 0.5) for the defined value of f(1).

Key Formula

lim⁡x→af(x)=Lbutf(a)≠L   or   f(a) is undefined\lim_{x \to a} f(x) = L \quad \text{but} \quad f(a) \neq L \;\text{ or }\; f(a) \text{ is undefined}
Where:
  • aa = The x-value where the discontinuity occurs
  • LL = The finite limit of f(x) as x approaches a
  • f(a)f(a) = The actual value of the function at x = a (which either differs from L or does not exist)

Worked Example

Problem: Determine whether the function f(x) = (x² − 4)/(x − 2) has a removable discontinuity, and if so, find the location of the hole.
Step 1: Check where the function is undefined. The denominator is zero when x = 2, so f(2) is undefined.
x−2=0  ⟹  x=2x - 2 = 0 \implies x = 2
Step 2: Factor the numerator and simplify the expression.
f(x)=x2−4x−2=(x−2)(x+2)x−2=x+2(x≠2)f(x) = \frac{x^2 - 4}{x - 2} = \frac{(x-2)(x+2)}{x-2} = x + 2 \quad (x \neq 2)
Step 3: Compute the limit as x approaches 2 using the simplified form.
lim⁡x→2f(x)=lim⁡x→2(x+2)=4\lim_{x \to 2} f(x) = \lim_{x \to 2} (x + 2) = 4
Step 4: Since the limit exists and equals 4, but f(2) is undefined, the discontinuity at x = 2 is removable. The hole is at the point (2, 4).
lim⁡x→2f(x)=4butf(2) is undefined\lim_{x \to 2} f(x) = 4 \quad \text{but} \quad f(2) \text{ is undefined}
Answer: Yes, there is a removable discontinuity (hole) at the point (2, 4). You can remove it by defining f(2) = 4.

Another Example

This example shows a case where the function IS defined at the point but its value doesn't match the limit — unlike the first example where the function was completely undefined. Both situations count as removable discontinuities.

Problem: Consider the piecewise function g(x) = { x + 3 for x ≠ 1, and g(1) = 10 }. Show that g has a removable discontinuity at x = 1.
Step 1: Compute the limit as x approaches 1. For all x ≠ 1, g(x) = x + 3.
lim⁡x→1g(x)=1+3=4\lim_{x \to 1} g(x) = 1 + 3 = 4
Step 2: Evaluate the function at x = 1. The piecewise definition gives g(1) = 10.
g(1)=10g(1) = 10
Step 3: Compare the limit to the function value. The limit exists and is finite, but it does not equal g(1).
lim⁡x→1g(x)=4≠10=g(1)\lim_{x \to 1} g(x) = 4 \neq 10 = g(1)
Step 4: Because the limit exists but disagrees with the function value, this is a removable discontinuity. Redefining g(1) = 4 would remove it.
Answer: g has a removable discontinuity at x = 1. The hole can be repaired by redefining g(1) = 4.

Frequently Asked Questions

What is the difference between a removable and non-removable discontinuity?
A removable discontinuity (hole) occurs when the limit exists at the point but the function value is missing or wrong — you can fix it by redefining a single point. A non-removable discontinuity cannot be repaired this way. Jump discontinuities have different left and right limits, while infinite discontinuities (vertical asymptotes) have limits that blow up to infinity. In both non-removable cases, no single-point redefinition can make the function continuous.
How do you find a removable discontinuity algebraically?
For rational functions, factor both the numerator and denominator completely. Any factor that cancels from both top and bottom produces a removable discontinuity at the x-value that makes that factor zero. The y-coordinate of the hole is found by substituting that x-value into the simplified expression. If a factor in the denominator does NOT cancel, it produces a vertical asymptote instead.
Is a removable discontinuity the same as a hole?
Yes. The terms are used interchangeably. A hole on a graph is the visual representation of a removable discontinuity: an open circle at a single missing or mismatched point where the curve would otherwise be connected.

Removable Discontinuity vs. Non-Removable Discontinuity

Removable DiscontinuityNon-Removable Discontinuity
DefinitionThe limit exists at the point, but the function is undefined or has a different value thereThe limit does not exist at the point (due to a jump, oscillation, or infinite behavior)
Graph appearanceA hole (open circle) in an otherwise continuous curveA jump between two pieces, a vertical asymptote, or wild oscillation
Can it be repaired?Yes — redefine the function at one point to equal the limitNo — no single-point redefinition can make the function continuous
Algebraic clue (rational functions)A common factor cancels from numerator and denominatorA denominator factor does NOT cancel (vertical asymptote) or the function has a piecewise jump

Why It Matters

Removable discontinuities appear frequently in precalculus and calculus when you simplify rational expressions or evaluate limits. Recognizing a hole lets you compute limits by cancellation — a technique central to finding derivatives from the definition. They also matter in real-world modeling: if a function has a removable discontinuity, it means the underlying pattern is well-behaved and only the definition needs a small fix.

Common Mistakes

Mistake: Concluding that because the simplified form gives a value at x = a, the original function is defined there.
Correction: The simplified form only tells you the limit. You must check the original function to see whether f(a) actually exists. For example, (x² − 4)/(x − 2) simplifies to x + 2, but the original function is still undefined at x = 2.
Mistake: Confusing a removable discontinuity (hole) with a vertical asymptote because both involve the denominator being zero.
Correction: A zero in the denominator that cancels with a factor in the numerator creates a hole. A zero that does NOT cancel creates a vertical asymptote. Always factor completely before deciding which type of discontinuity you have.

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