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Removing
Discontinuities

Learn how to identify and remove discontinuities by redefining a function at a single point: turning a "hole" into a continuous curve.

AP Calculus BC

What Is a Discontinuity?

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

1
\( f(a) \) is defined: the function has a value at \( x = a \).
2
\( \lim_{x \to a} f(x) \) exists: the left- and right-hand 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 does not equal the limit. Appears as a "hole" in the graph.
Jump
Left- and right-hand limits both exist but are not equal. The graph "jumps" at \( x = a \).
Infinite
At least one side tends to \( \pm\infty \). The graph has a vertical asymptote at \( x = a \).

Visualizing a Removable Discontinuity

x y 2 4 hole at (2, 4)

Why Removing Them Matters

Many core calculus theorems: the Intermediate Value Theorem, the Mean Value Theorem, and the Fundamental Theorem of Calculus: require continuity on an interval. A single removable discontinuity can invalidate their conclusions.

Derivatives
Differentiability implies continuity. A hole means the derivative is undefined there.
Integrals
Removing the hole lets us apply antiderivative techniques without worrying about the point of discontinuity.

The Removal Theorem

The key insight: the limit already tells us what the value should be. We simply assign it.

Before and After Removal

Before: hole at x = 2 2 → After: continuous 2

Removable vs Non-Removable

Removable Non-Removable
Two-sided limit Exists (finite) DNE or infinite
Graph appearance Hole (open dot) Jump or vertical asymptote
Fix by redefining? Yes No
Common source Factorable \( \frac{0}{0} \) Piecewise jumps, \( \frac{k}{0} \)

How to Remove a Discontinuity

1
Identify the trouble point \( x = a \) where \( f \) is undefined or disagrees with the limit.
2
Simplify or factor the expression to cancel the problematic factor (e.g., common \( (x - a) \) terms).
3
Compute the limit: \( L = \displaystyle\lim_{x \to a} f(x) \) using the simplified form.
4
Redefine: Set \( g(a) = L \) while keeping \( g(x) = f(x) \) for all other \( x \).
5
Verify that all three continuity conditions now hold at \( x = a \).

Example 1: \( \dfrac{x^2 - 9}{x - 3} \)

Find and remove the discontinuity of \( f(x) = \dfrac{x^2 - 9}{x - 3} \).

1
Trouble point: \( x = 3 \) gives \( \frac{0}{0} \).
2
Factor: \( \dfrac{(x-3)(x+3)}{x-3} = x + 3 \) for \( x \neq 3 \).
3
Limit: \( \displaystyle\lim_{x \to 3}(x+3) = \) \( 6 \).
4
Redefine: \( g(x) = \begin{cases} f(x), & x \neq 3 \\ 6, & x = 3 \end{cases} \)

Example 2: \( \dfrac{\sin x}{x} \)

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

x y 1 hole

Example 3: Piecewise Function

Consider the piecewise function:

1
Compute the limit: \( \displaystyle\lim_{x \to 2}(x^2 + 1) = 4 + 1 = \) \( 5 \).
2
Compare: \( f(2) = 10 \neq 5 \). The function value disagrees with the limit.
3
Redefine: Change \( f(2) \) from \(10\) to \(5\), making \( g(2) = 5 \).

Example 4: \( \dfrac{x^3 - 8}{x^2 - 4} \)

Remove the discontinuity of \( f(x) = \dfrac{x^3 - 8}{x^2 - 4} \) at \( x = 2 \).

1
Factor numerator: \( x^3 - 8 = (x - 2)(x^2 + 2x + 4) \).
2
Factor denominator: \( x^2 - 4 = (x - 2)(x + 2) \).
3
Cancel: \( \dfrac{(x-2)(x^2+2x+4)}{(x-2)(x+2)} = \dfrac{x^2+2x+4}{x+2} \) for \( x \neq 2 \).
4
Evaluate: \( \lim_{x \to 2} \dfrac{x^2+2x+4}{x+2} = \dfrac{4+4+4}{4} = \) \( 3 \). Set \( g(2) = 3 \).

When Removal Fails

If the two-sided limit at \( x = a \) does not exist (or is infinite), the discontinuity is non-removable. No single value can "bridge the gap."

Jump Example
\( f(x) = \begin{cases} 1, & x < 0 \\ 3, & x \ge 0 \end{cases} \)
\( \lim_{x \to 0^-}f(x) = 1 \), \( \lim_{x \to 0^+}f(x) = 3 \). No single value works.
Infinite Example
\( f(x) = \dfrac{1}{x} \) at \( x = 0 \).
The limit does not exist (tends to \( \pm\infty \)).

Common Mistakes

✕ Forgetting to check the limit
Students cancel factors and plug in without confirming the two-sided limit actually exists and is finite.
✕ Trying to remove a jump
Jump discontinuities cannot be fixed by redefining a single point. The left and right limits must agree first.
✕ Confusing the original function with the simplified one
After canceling, the simplified expression equals \( f(x) \) everywhere except at the trouble point. They are not the same function.

Summary Checklist

Step Action Goal
1 Find where \( f \) is undefined or mismatched Identify \( x = a \)
2 Factor and simplify Cancel common factors
3 Evaluate the limit \( L \) Confirm limit is finite
4 Redefine \( f(a) = L \) Fill the hole
5 Verify three continuity conditions Confirm continuity

Question 1 of 5

What value should be assigned to make \( f(x) = \dfrac{x^2 - 25}{x - 5} \) continuous at \( x = 5 \)?

Question 2 of 5

Which type of discontinuity does \( f(x) = \dfrac{1}{(x-3)^2} \) have at \( x = 3 \)?

Question 3 of 5

For the piecewise function \( f(x) = \begin{cases} x^2 - 1, & x \neq 1 \\ k, & x = 1 \end{cases} \), what value of \( k \) removes the discontinuity?

Question 4 of 5

Remove the discontinuity of \( f(x) = \dfrac{x^3 - 27}{x - 3} \) at \( x = 3 \). What is the required function value?

Question 5 of 5

A function has a discontinuity at \( x = a \) where \( \displaystyle\lim_{x \to a^-} f(x) = 4 \) and \( \displaystyle\lim_{x \to a^+} f(x) = 7 \). Is this discontinuity removable?