AP Calculus AB and BC

Removable vs Infinite Discontinuity

Ask what the two-sided limit does. If it exists as a finite number, the discontinuity is removable: the graph has a hole, and redefining that one point repairs it. If the function grows without bound near the point, the discontinuity is infinite: the graph has a vertical asymptote, and no redefinition can fix it.

Removable

Use when: Both one-sided limits agree on a finite value, which is what happens when a factor cancels and no denominator factor is left over.

Infinite

Use when: At least one one-sided limit runs off to positive or negative infinity, which is what happens when a denominator factor survives cancelling.

Side by side

	Removable	Infinite
Two-sided limit	Exists and is finite	Does not exist; the function is unbounded
Graph shows	A hole	A vertical asymptote
Source in a rational function	A factor that cancels	A denominator factor that survives
Can be repaired	Yes, by redefining one value	No, no assigned value would work
Common trap	Missing it because the cancelled form looks defined	Assuming every denominator zero produces one

The two-sided limit decides it. A removable discontinuity has a perfectly good limit that the function value fails to match or fails to supply. An infinite discontinuity has no limit at all, because the function leaves every bound as the input approaches the point.

\frac{x^2 - 4}{x^2 - x - 2} = \frac{(x - 2)(x + 2)}{(x - 2)(x + 1)}

One function shows both kinds. The factor $x - 2$ cancels, so $x = 2$ is a hole where the limit equals $\frac{4}{3}$ . The factor $x + 1$ stays in the denominator, so $x = -1$ is a vertical asymptote. Factoring separates the two before you draw anything.

Only one of them is repairable

Defining the value at $x = 2$ to be $\frac{4}{3}$ makes the function continuous there, because the limit was already waiting for it. At $x = -1$ there is nothing to match: the function runs to $-\infty$ from the left and $+\infty$ from the right, so no single value closes the gap.

Frequently asked questions

Is a hole still a discontinuity after I cancel the factor?

Yes. Cancelling rewrites the formula, not the function, so the original expression is still undefined at that input until the value is redefined.

Does an infinite discontinuity mean the limit equals infinity?

It means the function is unbounded near the point. Writing the limit as $\infty$ records that behaviour, but the limit does not exist as a number.

How do I tell them apart quickly?

Factor the numerator and denominator. A factor that cancels leaves a removable hole; a denominator factor that survives gives an infinite discontinuity.

In the CED: Unit 1: Limits and Continuity