Topic 1.13 Removing Discontinuities: recognize a removable discontinuity and define or redefine the function value to make it continuous.

What this topic is asking

The College Board (Topic 1.13) wants you to remove a removable discontinuity by assigning the right value at the hole. The rule is simple: redefine $f(a)$ to equal $\lim_{x \to a} f(x)$. You should also know why jump and infinite discontinuities cannot be removed this way.

The rule for removing a hole

How to do it

First confirm the discontinuity is removable: simplify the function (factor and cancel) and check that the two-sided limit exists and is finite. Then compute that limit. Finally, declare the function value at the point equal to the limit. The result is a piecewise definition: the original formula for $x \neq a$, plus $f(a) = L$.

The piecewise form of the patched function

When you remove a discontinuity, the honest way to write the result is as a piecewise function: the original formula for $x \neq a$, together with the assigned value at $x = a$. For the worked example below, the repaired function is "$\frac{x^3 - 8}{x - 2}$ for $x \neq 2$, and $12$ at $x = 2$". This makes explicit that you have not changed the function anywhere except the one previously-missing point - every other value is identical to the original. The simplified polynomial $x^2 + 2x + 4$ happens to agree with this patched function everywhere, which is why people often just say "the continuous extension is $x^2 + 2x + 4$". Both descriptions are correct; the piecewise form emphasizes that removal is a surgical fix at a single point, while the simplified form emphasizes that the hole was an artefact of how the function was written.

Why jumps and asymptotes cannot be removed

A removal works only because the limit exists, giving one target height for the value. At a jump, the left and right limits differ, so no single value can match both, and the two-sided limit does not exist - the third continuity condition can never be met. At an infinite discontinuity, the function grows without bound, so there is no finite value to assign at all. These are sometimes called non-removable (or essential) discontinuities. The exam often asks you to decide whether a given discontinuity is removable before asking you to remove it, so the first move is always to test whether the two-sided limit exists and is finite; only then does assigning a value make sense.

Why this idea matters

Removing a discontinuity is the concrete expression of the difference between a function's value and its limit, the theme that runs through all of Unit 1. A removable discontinuity is the one case where the limit exists but the value is wrong or missing, and patching it is simply forcing the value to equal the limit. The same idea reappears constantly in later units: many functions are defined by limits or have natural continuous extensions, and recognizing when a $\frac{0}{0}$ formula actually represents a perfectly smooth curve with one cosmetic gap is a habit you will use well beyond limits. It also sharpens the three-part continuity test, because a removable discontinuity fails only the third condition (value equals limit) while satisfying the first two.