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AP® Calculus AB-BC

What is a Removable Discontinuity?: AP® Calculus AB-BC Review

What is a removable discontinuity

Ever stared at a smooth curve that suddenly has a tiny gap? That “missing pixel” is a key idea on the AP® Calculus AB–BC exams (LIM-2.C.1 & LIM-2.C.2). Understanding it can earn easy points on limit questions. This post explains what a removable discontinuity is, how to see it on a graph, and how to “patch the hole” so the function becomes continuous.

What Is a Removable Discontinuity?

A removable discontinuity is a point where the graph should have a value, yet the function is either undefined or defined at a different height. In plain words, the curve wants to be connected, but something is missing. AP® synonyms include “hole” and “point discontinuity.”

Example #1

Identify and label the removable discontinuity off(x)=\dfrac{x^{2}-1}{x-1}

Solution

  1. Factor numerator: x^{2}-1=(x-1)(x+1).
  2. Simplify: f(x)=\dfrac{(x-1)(x+1)}{x-1}=x+1,\quad x\neq1.
  3. The simplified rule is x+1, yet x=1 is excluded. Therefore, the limit as x\to1 equals \lim_{x\to1}(x+1)=2.
  4. The graph displays a hole at (1,2).

Therefore, x=1 is a removable discontinuity.

Why Do Discontinuities Appear?

Algebra often creates gaps when a factor cancels a zero denominator. The original fraction is undefined, yet after canceling, the new expression behaves nicely. Graphically, the removable discontinuity graph shows a single open circle, not a jump or vertical asymptote.

Example #2

Show that g(x)=\dfrac{x^{2}-9}{x-3} has a removable discontinuity and find its location.

Solution

  1. Factor top: x^{2}-9=(x-3)(x+3).
  2. Simplify: g(x)=x+3,\quad x\neq3.
  3. The limit exists: \lim_{x\to3}(x+3)=6.
  4. Because g(3) is undefined, a hole occurs at (3,6).

How to Spot a Removable Discontinuity on a Graph

On a coordinate plane, a removable discontinuity appears as an open circle sitting on an otherwise smooth curve. However, a jump discontinuity has two distinct heights, and an infinite discontinuity shoots to ±∞.

Use this checklist on test day:

  • Are the left-hand and right-hand limits equal?
  • Is the function value either missing or different?
  • Does the graph stay finite near the point?

Example #3

A graph shows an open circle at (–2, 3) but otherwise follows the line y = –x + 1. Determine an equation that matches the graph.

Solution

  1. Write the rule of the line: y=-x+1.
  2. Introduce a factor causing the hole at x = –2: multiply by \dfrac{x+2}{x+2}.
  3. Equation: h(x)=\dfrac{(-x+1)(x+2)}{x+2}.
  4. The hole occurs where x+2=0\Rightarrow x=-2. The limit is h(-2)= -(-2)+1=3, matching the graph.

Removing the Discontinuity Algebraically

The AP® phrase “1.13 removing discontinuities” describes the simple fix:

  1. Factor numerator and denominator.
  2. Cancel the common factor.
  3. Redefine the function value at the troublesome x-value.

Example #4

Remove the discontinuity from p(x)=\dfrac{x^{2}-4x+4}{x-2} and write a new function.

Solution

  1. Factor: x^{2}-4x+4=(x-2)^{2}.
  2. Simplify: p(x)=x-2,\quad x\neq2.
  3. Limit: \lim_{x\to2}(x-2)=0.
  4. Define the patched function
P(x)=\begin{cases}\dfrac{x^{2}-4x+4}{x-2}, & x\neq2,\\[6pt]0, & x=2.\end{cases}

Now P(x) is continuous everywhere, satisfying LIM-2.C.1.

Piecewise-Defined Functions and Boundary Points

Continuity at a boundary requires the left-hand and right-hand limits to match the chosen function value (LIM-2.C.2). Therefore, limits are tested from both sides.

Example #5

Find k so that

f(x)=\begin{cases}kx+1, & x<4,\\[6pt]x^{2}-7, & x\ge4\end{cases}

is continuous at x=4.

Solution

  1. Left-hand limit: \lim_{x\to4^{-}}(kx+1)=4k+1.
  2. Right-hand limit & value: \lim_{x\to4^{+}}(x^{2}-7)=4^{2}-7=9.
  3. Set equal: 4k+1=9\Rightarrow4k=8\Rightarrow k=2.

Therefore, with k=2, the piecewise function is continuous. This aligns with “graphing a piecewise defined function” techniques.

Quick Reference Chart

TermDefinition
LimitExpected y-value as x approaches c
Removable DiscontinuityA hole where the limit exists but f(c) is wrong or missing
Continuous FunctionNo breaks; limit equals function value at every point
Boundary Pointx-value where pieces of a piecewise rule meet
Piecewise FunctionFunction defined by different rules on different intervals

AP® Exam Tips and Common Pitfalls

  • Always show factoring before canceling; scorers look for that step.
  • Never divide by zero when simplifying.
  • After canceling, explicitly state the value used to patch the hole.
  • Clearly write one-sided limits when verifying continuity at a boundary.
  • Label holes on sketches; many free-response rubrics allot a point for this detail.

Conclusion

Removable discontinuities are friendly hurdles: the limit exists, and a simple redefinition repairs the graph. Mastering this skill boosts confidence with limits, piecewise functions, and graph interpretation. Keep practicing “What is a removable discontinuity?” on both paper and calculator views. That small hole can turn into an easy point on the AP® test—and a smoother path through calculus.

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