01 / 21

Removing
Discontinuities

Learn how to identify removable discontinuities and redefine functions to make them continuous at a point.

AP Calculus AB

What Is a Discontinuity?

A function \( f(x) \) is continuous at \( x = a \) when three conditions hold simultaneously. If any one fails, the function has a discontinuity at that point.

1
\( f(a) \) is defined: the function has a value at the point.
2
\( \lim_{x \to a} f(x) \) exists: the left and right limits agree.
3
\( \lim_{x \to a} f(x) = f(a) \): the limit equals the function value.

Types of Discontinuities

Removable
The limit exists but \( f(a) \) is either undefined or doesn't equal the limit. A "hole" in the graph.
Jump
Left-hand and right-hand limits both exist but are not equal. The graph "jumps" between values.
Infinite
The function approaches \( \pm\infty \) as \( x \to a \). A vertical asymptote appears.

Visualizing a Removable Discontinuity

The Core Idea: Filling the Hole

If \( f \) has a removable discontinuity at \( x = a \), we can define a new function \( g \) that is continuous at \( a \):

In other words: evaluate the limit, then assign that value to the function at the problem point. The "hole" gets filled, and \( g \) is continuous everywhere \( f \) was, plus at \( x = a \).

Formal Theorem

The discontinuity is removed by defining (or redefining) \( f(a) = L \). After this change, all three continuity conditions are satisfied.

Removable vs Non-Removable

Removable Non-Removable
Limit at \( x = a \) Exists and is finite DNE or infinite
Graph appearance Hole (open circle) Jump or vertical asymptote
Algebraic sign Common factor cancels Factor does not cancel
Can be fixed? Yes: redefine \( f(a) \) No
Example \( \frac{x^2-4}{x-2} \) at \( x=2 \) \( \frac{1}{x-2} \) at \( x=2 \)

How to Remove a Discontinuity

1
Identify the point: Find where \( f(x) \) is undefined or where \( f(a) \neq \lim_{x \to a} f(x) \).
2
Factor and simplify: Factor numerator and denominator, then cancel the common factor \( (x - a) \).
3
Evaluate the limit: Substitute \( x = a \) into the simplified expression to find \( L = \lim_{x \to a} f(x) \).
4
Redefine: Set \( f(a) = L \). The function is now continuous at \( x = a \).

Example 1: Factoring

Remove the discontinuity of \( f(x) = \dfrac{x^2 - 9}{x - 3} \) at \( x = 3 \).

1
Factor: \( x^2 - 9 = (x-3)(x+3) \)
2
Cancel: \( \dfrac{(x-3)(x+3)}{x-3} = x + 3 \) for \( x \neq 3 \)
3
Evaluate: \( \displaystyle\lim_{x \to 3}(x+3) = 6 \)
4
Redefine: Set \( f(3) = 6 \). Now \( f \) is continuous at \( x = 3 \).

Example 1: Before and After

Example 2: Trig Limits

Remove the discontinuity of \( f(x) = \dfrac{\sin x}{x} \) at \( x = 0 \).

1
Problem: \( f(0) = \frac{\sin 0}{0} = \frac{0}{0} \), which is undefined.
2
Known limit: \( \displaystyle\lim_{x \to 0} \frac{\sin x}{x} = 1 \) (the Squeeze Theorem result).
3
Redefine: Set \( f(0) = 1 \). The discontinuity is removed.

Example 3: Piecewise Functions

Find the value of \( k \) that removes the discontinuity:

1
Factor: \( \frac{x^2-4}{x-2} = \frac{(x-2)(x+2)}{x-2} = x+2 \) for \( x \neq 2 \).
2
Limit: \( \displaystyle\lim_{x \to 2}(x+2) = 4 \).
3
Set \( k = 4 \): Now \( f(2) = 4 = \lim_{x \to 2} f(x) \), so \( f \) is continuous.

Example 3: Choosing \( k \)

Beyond Factoring

Factoring handles polynomials, but other techniques may be needed:

Conjugates
Multiply by the conjugate for expressions with radicals: \( \frac{\sqrt{x}-2}{x-4} \).
Known Limits
Use standard results: \( \frac{\sin x}{x} \to 1 \), \( \frac{1-\cos x}{x} \to 0 \).
L'Hopital's Rule
For \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) forms, differentiate numerator and denominator.

Common Mistakes

✕ Direct Substitution
Plugging \( x = a \) into the original before simplifying gives \( \frac{0}{0} \), not a real answer.
✕ Forgetting the Domain
After canceling, stating the simplified form works for all \( x \) without noting the restriction.
✕ Removing Non-Removables
Trying to "fill" a vertical asymptote or jump: these discontinuities cannot be removed by redefining one point.

Summary Checklist

Step Action Result
1 Find where \( f \) is undefined Identify \( x = a \)
2 Compute \( \lim_{x\to a} f(x) \) Get \( L \) (finite?)
3 Check: Is \( L \) finite? Yes: removable
4 Redefine \( f(a) = L \) Continuous at \( a \)

Question 1 of 5

What value of \( f(3) \) removes the discontinuity of \( f(x) = \dfrac{x^2 - 5x + 6}{x - 3} \)?

Question 2 of 5

Which type of discontinuity does \( f(x) = \dfrac{1}{x - 5} \) have at \( x = 5 \)?

Question 3 of 5

For \( g(x) = \dfrac{x^2 - 1}{x + 1} \), what value removes the discontinuity at \( x = -1 \)?

Question 4 of 5

Find \( k \) so that \( f(x) = \begin{cases} \frac{\sin(2x)}{x} & x \neq 0 \\ k & x = 0 \end{cases} \) is continuous at \( x = 0 \).

Question 5 of 5

True or false: If \( \displaystyle\lim_{x \to a} f(x) \) exists, then \( f \) must be continuous at \( x = a \).