Calculus I · Limits and Continuity · lesson

How to Repair a Removable Discontinuity

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)= \begin{cases} \dfrac{x^2-4}{x-2},&x\ne2,\\ c,&x=2. \end{cases}

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)= \begin{cases} \dfrac{x^3-27}{x-3},&x\ne3,\\ c,&x=3 \end{cases}

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.

After the explanation

Use the section idea

Reading lens

Do the limit, the function value, and the surrounding domain fit together at the point or across the interval?

Mental model

Continuity is a three-part agreement: the value exists, the two-sided limit exists, and those two quantities are equal.

Decision

At a point, test the three conditions in order; on an interval, check the domain and endpoints before invoking any continuity theorem.

Common trap

A sign change supports the Intermediate Value Theorem only when continuity holds on the entire closed interval, and it does not prove uniqueness.

Check yourself

You are ready to continue when you can classify a break, decide whether one value can repair it, and state every IVT hypothesis aloud.

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.