What is a removable discontinuity, and how can it be recognized?
Quick answer
A removable discontinuity at x = a is a hole or missing/mismatched point where the two-sided limit exists and is finite, even though the function’s value there does not match that limit or is undefined. To recognize one, find the limit from nearby x-values and check the original function’s domain. If factoring and cancellation make the nearby expression approach a finite number, that number is the limit—not automatically the function’s value at the hole. [S1]
The question, the details, the next step.
I keep mixing up a hole in a graph with a point where the function just gets very large. What makes a discontinuity removable?
The key question is whether the function’s two-sided limit at the problem x-value exists and is a finite real number. If it does, but the function is undefined there or its actual value differs from that limit, the discontinuity is removable: changing the value at that one input would make the function continuous there. [S1]
So don’t decide from a graph’s appearance alone. Identify the x-value, ask what the function approaches from both sides, and separately check whether the original function actually has a value at that input. A finite limit describes nearby behavior; it does not, by itself, fill in a missing point.
Can you show that with the fraction (x² − 4)/(x − 2)? How do I find where the hole is and what value would fill it?
Start with the original expression. At x = 2, its denominator is zero, so the original function is undefined there. Factor the numerator:
- x² − 4 = (x − 2)(x + 2).
- For x ≠ 2, cancel the common factor to get x + 2.
- As x approaches 2, x + 2 approaches 4.
Thus the hole is at the point (2, 4): the x-coordinate comes from the excluded input, and the y-coordinate is the finite limit. Defining the function’s value at x = 2 to be 4 would fill the hole. The cancellation describes values near 2; it does not change the original function’s domain or assign it a value at 2. [S1]
What should I check when I factor and cancel? I’m worried that canceling something means I’ve fixed the original function.
Keep two records: the simplified expression for nearby inputs and the restrictions from the original formula. A factor can cancel algebraically where it is nonzero, but if that factor made the original denominator zero at a particular input, that input remains excluded from the original function. [S1]
For a candidate x = a, use this short check:
- Substitute into the original denominator to identify any excluded input.
- Factor and cancel only for nearby inputs where cancellation is valid.
- Evaluate the simplified expression’s limit as x approaches a.
- Confirm that the two-sided limit is finite.
- Compare that limit with the original function’s value at a, if one is defined.
A finite limit plus a missing or mismatched value is the removable case. If the two-sided limit does not exist or is not finite, this test does not establish a removable discontinuity. [S1]
Can you give me a reusable way to write the conclusion, so I don’t report the limit as if it were already the function value?
Write the limit and the function value as separate statements. For the example, a clear conclusion is:
“Since the original denominator is zero at x = 2, the function is undefined there. For x ≠ 2, it simplifies to x + 2, so the limit as x approaches 2 is 4. The graph has a removable discontinuity, or hole, at (2, 4). Assigning f(2) = 4 would fill the hole.”
That wording distinguishes what the function does nearby from what it is actually defined to be at the point. More generally, report the hole as (a, L), where a is the excluded input and L is the finite limit, and say that setting f(a) = L would fill it. [S1]
A practical checklist
Find the candidate x-value and check it in the original denominator or domain.
Factor the expression and cancel only when describing values away from the excluded input.
Calculate the two-sided limit at the candidate input.
Confirm the limit is finite, then compare it with the function’s actual value there.
State the hole as (excluded input, limit) and distinguish it from the original function value.
A few more useful answers
Does canceling a common factor make the original function defined at the hole?
No. Cancellation gives an equivalent expression only for inputs where the canceled factor is nonzero. Check the original domain: the excluded input stays excluded unless the function is explicitly redefined there. [S1]
What value would fill a removable discontinuity?
The finite two-sided limit at the hole’s x-coordinate. In the worked example, that limit is 4, so assigning f(2) = 4 fills the hole. [S1]
Does a finite one-sided limit prove a discontinuity is removable?
Not by itself. The criterion requires the two-sided limit to exist and be finite; check that the function approaches the same value from both sides. [S1]
Sources
References used for this guide.
- [S1] OpenStax Calculus Volume 1, Section 2.4: Continuity and Discontinuities openstax.org · Checked