Calculus I · Limits and Continuity · lesson

How to Repair a Removable Discontinuity

Learn How to Repair a Removable Discontinuity with plain-language explanations, guided examples, worked homework methods, interactive checks, and exam-style pra

Course progress37 of 47 core pages37 of 47

Where this chapter fits

Chapter 5: Continuity

Connect limits to function values, classify discontinuities, repair piecewise definitions, and use the Intermediate Value Theorem.

Reading lens: Do the limit, the function value, and the surrounding domain fit together at the point or across the interval? Keep that question in view while reading How to Repair a Removable Discontinuity; the worked mathematics is evidence for the idea, not a substitute for it.

This page connects Types of Discontinuity to Choosing Parameters for Piecewise Continuity. Read the explanation first, predict each example’s next move, and only then compare the written solution.

Concept

Learning objectives

Choose a missing function value so that a function becomes continuous at a point.

Repairing Removable Discontinuities

If a finite limit exists at $a$ , define the missing or incorrect function value to equal that limit.

Guided walkthrough

Fill the hole with the approached value

Let

f(x)= \left\{\matrix{ \dfrac{x^2-4}{x-2},\quad x\ne2,\quad c,\quad x=2. }\right.

Find $c$ so that $f$ is continuous at $2$ .

Show worked solution

For $x\ne2$ ,

\frac{x^2-4}{x-2}=x+2.

Therefore,

\lim_{x\to2}f(x)=4.

Continuity requires

f(2)=\lim_{x\to2}f(x).

Since $f(2)=c$ , choose

\boxed{c=4}.

Interactive checkcontinuity-flow-01

Choose $c$ so $f(x)=\frac{x^2-16}{x-4}$ for $x\ne4$ , and $f(4)=c$ , is continuous.

Your work stays on this device. No account or AI grader is used.

Show hint

Find the limit at $4$ after factoring.

Attempt once to unlock the solution

Submit an answer first. The hint is available now.

Worked example

Repair a more complicated hole

Choose $c$ so that

g(x)= \left\{\matrix{ \dfrac{x^3-27}{x-3},\quad x\ne3,\quad c,\quad x=3 }\right.

is continuous at $3$ .

Show worked solution

Factor the difference of cubes:

x^3-27=(x-3)(x^2+3x+9).

Thus, for $x\ne3$ ,

g(x)=x^2+3x+9.

Take the limit:

3^2+3(3)+9=9+9+9=27.

Therefore,

\boxed{c=27}.

Common mistake

Only removable discontinuities can be repaired by changing one function value. A jump or vertical asymptote reflects the behavior of infinitely many nearby points, not one misplaced dot.

Source & rights

Original instruction with traceable references.

The exposition is original. No Active Calculus exercise is reproduced verbatim. Public-domain examples were modernized and recomposed when used as inspiration.

The verified handoff declares original composition and requires owner provenance review. BetterGrades-original material remains separate from public-domain references; no source textbook PDF is published here.

Chapter 5: Continuity

Connect limits to function values, classify discontinuities, repair piecewise definitions, and use the Intermediate Value Theorem.

Repairing Removable Discontinuities

If a finite limit exists at

a

, define the missing or incorrect function value to equal that limit.

Guided walkthrough

Fill the hole with the approached value

Let

f(x)= \left\{\matrix{ \dfrac{x^2-4}{x-2},\quad x\ne2,\quad c,\quad x=2. }\right.

Find $c$ so that $f$ is continuous at $2$ .

Show worked solution

For $x\ne2$ ,

\frac{x^2-4}{x-2}=x+2.

Therefore,

\lim_{x\to2}f(x)=4.

Continuity requires

f(2)=\lim_{x\to2}f(x).

Since $f(2)=c$ , choose

\boxed{c=4}.

Interactive checkcontinuity-flow-01

Choose $c$ so $f(x)=\frac{x^2-16}{x-4}$ for $x\ne4$ , and $f(4)=c$ , is continuous.

Your work stays on this device. No account or AI grader is used.

Show hint

Find the limit at $4$ after factoring.

Attempt once to unlock the solution

Submit an answer first. The hint is available now.

Worked example

Repair a more complicated hole

Choose $c$ so that

g(x)= \left\{\matrix{ \dfrac{x^3-27}{x-3},\quad x\ne3,\quad c,\quad x=3 }\right.

is continuous at $3$ .

Show worked solution

Factor the difference of cubes:

x^3-27=(x-3)(x^2+3x+9).

Thus, for $x\ne3$ ,

g(x)=x^2+3x+9.

Take the limit:

3^2+3(3)+9=9+9+9=27.

Therefore,

\boxed{c=27}.

Common mistake

Only removable discontinuities can be repaired by changing one function value. A jump or vertical asymptote reflects the behavior of infinitely many nearby points, not one misplaced dot.

Source & rights

Original instruction with traceable references.

The exposition is original. No Active Calculus exercise is reproduced verbatim. Public-domain examples were modernized and recomposed when used as inspiration.

Chapter 5: Continuity

Repairing Removable Discontinuities

Related resources

Original instruction with traceable references.

Chapter 5: Continuity

Repairing Removable Discontinuities

Related resources

Original instruction with traceable references.