Exploring Types of
Discontinuities
Understand where functions break, how to classify removable, jump, and infinite discontinuities, and what they mean for limits and continuity.
AP Calculus AB
What Does Continuity Mean?
A function \( f(x) \) is continuous at a point \( x = a \) if you can draw through that point without lifting your pen. Formally, three conditions must hold simultaneously:
1
\( f(a) \) is defined: the function has a value at that point.
2
\( \lim_{x \to a} f(x) \) exists: the left-hand and right-hand limits agree.
3
\( \lim_{x \to a} f(x) = f(a) \): the limit equals the function value.
Why Discontinuities Matter
Discontinuities are not just abstract curiosities: they have real consequences throughout calculus and applied mathematics.
Limits
Discontinuities reveal where direct substitution fails and special limit techniques are needed.
Derivatives
A function must be continuous at a point to be differentiable there. Discontinuities block differentiation.
Integrals
The Fundamental Theorem of Calculus requires continuity on the interval. Discontinuities may require splitting integrals.
The Three Types of Discontinuity
Removable
A "hole" in the graph. The limit exists, but \( f(a) \) is missing or wrong. You can "fill" it.
Jump
The function "jumps" from one value to another. Left and right limits exist but are not equal.
Infinite
The function blows up to \( \pm\infty \). At least one side diverges: a vertical asymptote.
Removable Discontinuity
Jump Discontinuity
Infinite Discontinuity
Formal Definitions
R
Removable: \( \lim_{x \to a} f(x) = L \) exists (finite), but \( f(a) \neq L \) or \( f(a) \) is undefined.
J
Jump: \( \lim_{x \to a^-} f(x) = L_1 \) and \( \lim_{x \to a^+} f(x) = L_2 \) both exist, but \( L_1 \neq L_2 \).
I
Infinite: \( \lim_{x \to a^-} f(x) = \pm\infty \) or \( \lim_{x \to a^+} f(x) = \pm\infty \) (or both). Vertical asymptote at \( x = a \).
Removable vs Non-Removable
| Removable | Non-Removable | |
|---|---|---|
| Two-sided limit | Exists (finite) | DNE or infinite |
| One-sided limits | Equal each other | Unequal or divergent |
| Graph feature | Hole (open circle) | Jump or asymptote |
| Can be "fixed"? | Yes: redefine \( f(a) = L \) | No: structural break |
| Algebraic clue | Common factor cancels | No cancellation possible |
How to Classify a Discontinuity
1
Find the trouble point. Where is \( f(x) \) undefined or where does the formula change?
2
Compute the left-hand limit \( \lim_{x \to a^-} f(x) \). Is it finite or infinite?
3
Compute the right-hand limit \( \lim_{x \to a^+} f(x) \). Is it finite or infinite?
4
Compare: If both are finite and equal: removable. If both finite but unequal: jump. If either is infinite: infinite.
5
Check \( f(a) \). For removable: is the limit equal to \( f(a) \)? If yes, it is actually continuous.
Example 1: Removable
Classify the discontinuity of \( f(x) = \dfrac{x^2 - 9}{x - 3} \) at \( x = 3 \).
1
Trouble point: \( x = 3 \) makes the denominator zero. \( f(3) \) is undefined.
2
Factor: \( \dfrac{x^2 - 9}{x - 3} = \dfrac{(x-3)(x+3)}{x-3} = x + 3 \) for \( x \neq 3 \).
3
Compute limit: \( \lim_{x \to 3} (x + 3) = \) \( 6 \). Both sides agree.
4
Conclusion: Limit exists and is finite, but \( f(3) \) is undefined. Removable discontinuity.
Example 2: Jump
Classify the discontinuity of \( f(x) = \begin{cases} x + 1 & x < 2 \\ 2x - 3 & x \geq 2 \end{cases} \) at \( x = 2 \).
1
Left-hand limit: \( \lim_{x \to 2^-} (x + 1) = \) \( 3 \).
2
Right-hand limit: \( \lim_{x \to 2^+} (2x - 3) = \) \( 1 \).
3
Compare: \( 3 \) \( \neq \) \( 1 \). The one-sided limits disagree.
4
Conclusion: Jump discontinuity at \( x = 2 \). The function jumps from \( 3 \) to \( 1 \).
Example 3: Infinite
Classify the discontinuity of \( f(x) = \dfrac{1}{(x - 1)^2} \) at \( x = 1 \).
Algebraic Detection for Rational Functions
For a rational function \( f(x) = \dfrac{P(x)}{Q(x)} \), factor both numerator and denominator completely, then examine common factors:
Common factor cancels
If \( (x - a) \) appears in both \( P(x) \) and \( Q(x) \) and cancels completely, there is a removable discontinuity at \( x = a \).
Factor remains in denominator
If after cancellation \( (x - a) \) still appears in the denominator, there is an infinite discontinuity (vertical asymptote) at \( x = a \).
Common Mistakes
✕ Confusing Hole with Asymptote
A cancelled factor creates a hole (removable), not an asymptote. Only factors remaining in the denominator create vertical asymptotes.
✕ Forgetting to Check \( f(a) \)
Even when the limit exists, if \( f(a) = L \), the function is actually continuous. Always verify whether the value matches the limit.
✕ Calling Jump "Removable"
If the left and right limits are different finite values, no redefinition of \( f(a) \) can fix it. The two-sided limit does not exist.
Summary Comparison
| Removable | Jump | Infinite | |
|---|---|---|---|
| \( \lim_{x \to a} f(x) \) | Exists (= L) | DNE | DNE (\( \pm\infty \)) |
| One-sided limits | Both = \( L \) | \( L_1 \neq L_2 \), both finite | At least one \( = \pm\infty \) |
| Graph clue | Hole (open circle) | Step / break | Vertical asymptote |
| Fixable? | Yes | No | No |
| Typical source | Cancelled factor | Piecewise function | Zero in denominator |
Question 1 of 5
What type of discontinuity does \( f(x) = \dfrac{x^2 - 1}{x - 1} \) have at \( x = 1 \)?
Factor: \( \frac{(x-1)(x+1)}{x-1} = x + 1 \) for \( x \neq 1 \). The limit is \( 2 \), but \( f(1) \) is undefined. Since the limit exists but the function value doesn't match, this is a removable discontinuity.
Question 2 of 5
If \( \lim_{x \to 4^-} f(x) = 7 \) and \( \lim_{x \to 4^+} f(x) = 3 \), the discontinuity at \( x = 4 \) is:
Both one-sided limits are finite but unequal (\( 7 \neq 3 \)). This is the defining feature of a jump discontinuity. The two-sided limit does not exist.
Question 3 of 5
The function \( g(x) = \dfrac{x + 5}{x^2 - 25} \) has a removable discontinuity at which value of \( x \)?
Factor: \( \frac{x+5}{(x+5)(x-5)} = \frac{1}{x-5} \) for \( x \neq -5 \). The factor \( (x + 5) \) cancels, creating a removable discontinuity at \( x = -5 \). At \( x = 5 \), the factor remains: that's an infinite discontinuity.
Question 4 of 5
A function has a vertical asymptote at \( x = 2 \). Which type of discontinuity occurs there?
A vertical asymptote means \( f(x) \to \pm\infty \) as \( x \to 2 \) from at least one side. This is the definition of an infinite discontinuity. The limit diverges, so it cannot be removable or a jump.
Question 5 of 5
Which condition, if satisfied, guarantees that a discontinuity at \( x = a \) is removable?
A discontinuity is removable if and only if the two-sided limit exists and is finite. Whether \( f(a) \) is defined or what its value is doesn't determine the type: it determines whether the point is actually discontinuous. The key criterion for "removable" is that the limit exists.