Removing Discontinuities — AP Calculus AB
1. What is Removing a Discontinuity? ★☆☆☆☆ ⏱ 2 min
Removing a discontinuity (also called filling a hole) is the process of redefining a function at a single point to make it continuous at that point, with no changes to the function anywhere else. This topic falls in Unit 1, which accounts for 10-12% of your total AP Calculus AB exam score, and appears in both multiple-choice and free-response questions.
2. Identifying Removable Discontinuities ★★☆☆☆ ⏱ 4 min
Before you can remove a discontinuity, you must first correctly identify it as removable, rather than non-removable. Recall the three required conditions for continuity at a point $x=a$:
- $f(a)$ is defined
- $\lim_{x \to a} f(x)$ exists
- $\lim_{x \to a} f(x) = f(a)$
A discontinuity is removable if only conditions 1 or 3 fail, but condition 2 still holds (the two-sided limit exists). If condition 2 fails, the discontinuity is non-removable and cannot be fixed by changing $f(a)$. Most removable discontinuities arise in rational functions when the numerator and denominator share a common linear factor $(x-a)$, creating a $0/0$ indeterminate form.
Exam tip: Always check every root of the denominator, not just the one that cancels. AP exam questions almost always include one non-removable discontinuity alongside a removable one to test if you check all discontinuities.
3. Removing Discontinuities by Factoring ★★☆☆☆ ⏱ 4 min
Once you have identified a removable discontinuity at $x=a$, removing it requires you to define or redefine $f(a)$ to equal the value of $\lim_{x \to a} f(x)$, which satisfies all three continuity conditions at $x=a$. For polynomial rational functions, the standard method to find this limit is factoring and canceling common factors.
Exam tip: Never skip explicitly stating that $f(a)$ equals the limit. AP graders will deduct points if you only simplify the function and do not explicitly define the new value of $f(a)$ to remove the discontinuity.
4. Removing Discontinuities by Rationalizing ★★★☆☆ ⏱ 4 min
Not all removable discontinuities come from factoring polynomial rational functions. Another common case is functions with radicals, where the function is undefined at $x=a$ (giving a $0/0$ indeterminate form) but the limit still exists. To evaluate this limit, we multiply both the numerator and denominator by the conjugate of the radical expression to eliminate the radical, which reveals a common $(x-a)$ factor that can be canceled.
Exam tip: Always multiply both the numerator and denominator by the conjugate. If you only multiply the numerator, you change the value of the expression, leading to an incorrect limit.
Common Pitfalls
Why: Students confuse the simplified function (defined at $a$) with the original function, which remains undefined at $a$ unless explicitly changed.
Why: Students are taught $0/0$ usually means removable, but rare cases can still have non-existent limits.
Why: Students rush to eliminate the radical and forget the multiplication rule for equality.
Why: Students develop a habit of canceling any matching factor, even when the numerator is not zero.
Why: $g(x)$ is often undefined at $a$ in this setup, so students incorrectly plug in $a$ to the simplified $g(x)$.