Types of Discontinuity

Learn to tell removable discontinuities (holes), jump discontinuities and infinite discontinuities apart, from graphs and from formulas.

Advanced30 min18 problems
Practice Problems

Definition

A function ff is continuous at x=ax = a when lim⁡x→af(x)=f(a)\lim_{x \to a} f(x) = f(a). If this fails, ff has a discontinuity at x=ax = a. How it fails tells you the type:
1. Removable discontinuity (hole) The limit exists and is a finite number LL, but f(a)f(a) is missing or f(a)≠Lf(a) \neq L. The graph has a single hole at (a,L)(a, L). You can remove it by redefining f(a)=Lf(a) = L.
2. Jump discontinuity Both one-sided limits are finite but different:
lim⁡x→a−f(x)≠lim⁡x→a+f(x)\lim_{x \to a^-} f(x) \neq \lim_{x \to a^+} f(x)
The graph jumps from one height to another. The jump size is ∣lim⁡x→a+f(x)−lim⁡x→a−f(x)∣\left| \lim_{x \to a^+} f(x) - \lim_{x \to a^-} f(x) \right|. No single value of f(a)f(a) can fix it.
3. Infinite discontinuity At least one one-sided limit is ∞\infty or −∞-\infty. The graph has a vertical asymptote x=ax = a.
From a formula (rational functions): factor the numerator and the denominator.
  • A factor (x−a)(x - a) that cancels completely gives a hole at x=ax = a.
  • A factor (x−a)(x - a) that stays in the denominator gives a vertical asymptote at x=ax = a.

Try it now

What type of discontinuity does f(x)=x2−16x−4f(x) = \frac{x^2 - 16}{x - 4} have at x=4x = 4?

Worked Examples

Classify the discontinuity of f(x)=x2−x−2x−2f(x) = \frac{x^2 - x - 2}{x - 2} at x=2x = 2 and remove it.
1

Check where ff is undefined

The denominator x−2x - 2 is 00 at x=2x = 2 → f(2)f(2) is undefined
2

Factor the numerator

x2−x−2=(x−2)(x+1)x^2 - x - 2 = (x - 2)(x + 1) → f(x)=(x−2)(x+1)x−2f(x) = \frac{(x - 2)(x + 1)}{x - 2}
3

Cancel the common factor

f(x)=x+1f(x) = x + 1 for every x≠2x \neq 2 → The factor (x−2)(x - 2) cancels completely
4

Find the limit

lim⁡x→2f(x)=2+1\lim_{x \to 2} f(x) = 2 + 1 → =3= 3, a finite limit
5

Classify and fix

Limit exists, f(2)f(2) does not: a hole at (2,3)(2, 3). Define f(2)=3f(2) = 3. → Removable discontinuity

Common Mistakes

Drawing a vertical asymptote at every zero of the denominator

Why it's wrong: A zero of the denominator can also be a hole. In x2−x−2x−2\frac{x^2 - x - 2}{x - 2} the factor (x−2)(x - 2) cancels, so there is no asymptote at x=2x = 2.

Correct: Factor and cancel first. Only factors left in the denominator give vertical asymptotes; cancelled factors give holes.

Calling a jump removable because both one-sided limits exist

Why it's wrong: Removable needs the two-sided limit, which exists only when both sides agree. At a jump they are different numbers.

Correct: Equal one-sided limits: removable (or continuous). Different finite one-sided limits: jump.

Using f(a)f(a) as the height of the hole

Why it's wrong: At a hole f(a)f(a) is missing or wrong. The hole sits at the height of the limit.

Correct: Find the yy-coordinate of a hole by substituting x=ax = a into the simplified function: the hole of x2−9x2−x−6\frac{x^2 - 9}{x^2 - x - 6} is at (3,65)\left(3, \frac{6}{5}\right).

Thinking a cancelled factor always means a hole, even when it appears twice in the denominator

Why it's wrong: In x−1(x−1)2\frac{x - 1}{(x - 1)^2} only one factor cancels; 1x−1\frac{1}{x - 1} is left, which still blows up at x=1x = 1.

Correct: Cancel as far as possible, then check: if (x−a)(x - a) is still in the denominator, it is an infinite discontinuity.

Interactive Visual

Function graph

-3-2-1012345-20246xy(2, 3)

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f(x) = (x² − x − 2)/(x − 2)

Table of values

x1.91.9922.012.1
y2.92.99undef.3.013.1

Function graph

-3-2-101230246xy

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g(x)
  • At x = 1: from the left and from the right g(x) → 1, so the limit is 1. g(1) = 4.

Table of values

x0.90.9911.011.1
y0.810.980141.02011.21

Function graph

-3-2-101230510xy

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h(x)
  • At x = 1: from the left h(x) → 3, from the right h(x) → 6. The values differ, so the limit does not exist (jump). h(1) = 6.

Table of values

x0.90.9911.011.1
y2.92.9966.02016.21

Function graph

-8-6-4-202468-505xy(3, 1.2)x = -2y = 1

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r(x) = (x² − 9)/(x² − x − 6)Asymptote

Table of values

x-2.1-2.01-1.99-1.92.993.01
y-9-99101111.20041.1996

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Interactive Grapher

-10-50510-10-50510xy

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y = 2x + 1

m=2, b=1

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Practice Problems

18 problems
Problem 1 of 18
Easy
What type of discontinuity does f(x)=x2−16x−4f(x) = \frac{x^2 - 16}{x - 4} have at x=4x = 4?

Why It Matters

Knowing the type of a discontinuity tells you how serious the break is and what you can do about it:
  • Holes are harmless gaps: the function is predictable around them, and one redefined point makes it continuous. Computer programs that crash on 00\frac{0}{0} are often fixed exactly this way.
  • Jumps describe sudden changes: parking fees, shipping rates, tax brackets and switches that turn on or off.
  • Infinite discontinuities warn that a quantity blows up: lens images moving to infinity, resonance in bridges, or a formula dividing by something close to zero.
In calculus, the type of discontinuity decides whether a limit exists, whether a function can be integrated over an interval, and how a graph must be drawn. Graphing rational functions correctly (holes versus asymptotes) is one of the most tested skills before calculus.

Real World Applications

Parking Fees (Jumps)

Parking garages charge by the started hour, so the price function is constant on each hour and jumps at every full hour.

Example:

A garage charges 4 dollars for up to 1 hour and 7 dollars for more than 1 hour up to 2 hours. At t=1t = 1 the price jumps by 3 dollars.

1Try It Yourself

The same garage charges 7 dollars for more than 1 hour up to 2 hours and 10 dollars for more than 2 hours up to 3 hours. C(t)C(t) is the price for tt hours.

What is the size of the jump in the price at t=2t = 2 hours (in dollars)?

Step 1: Write the mathematical expression

Subtract the left-hand limit from the right-hand limit:

Lenses and Images at Infinity

For a lens with focal length $10$ cm, an object at distance $u$ produces an image at distance $v(u) = \frac{10u}{u - 10}$. When the object reaches the focal point, the image moves off to infinity.

Example:

At u=15u = 15 cm the image is at v=1505=30v = \frac{150}{5} = 30 cm. At u=10.1u = 10.1 cm it is already at v=1010.1=1010v = \frac{101}{0.1} = 1010 cm.

2Try It Yourself

The image distance is v(u)=10uu−10v(u) = \frac{10u}{u - 10} (in cm).

At which object distance uu (in cm) does vv have an infinite discontinuity?

Step 1: Write the mathematical expression

Find where the denominator is zero and the numerator is not:

Fixing a Division by Zero in Software

Programs often compute an average rate of change $\frac{f(t) - f(a)}{t - a}$. At $t = a$ the formula gives $\frac{0}{0}$ and the program crashes, although the values nearby are perfectly smooth: a removable discontinuity.

Example:

For f(t)=t2f(t) = t^2 and a=3a = 3, the rate t2−9t−3\frac{t^2 - 9}{t - 3} simplifies to t+3t + 3, so the programmer stores the value 66 for t=3t = 3.

3Try It Yourself

A program computes R(t)=t2−16t−4R(t) = \frac{t^2 - 16}{t - 4} and crashes at t=4t = 4.

Which value should the programmer assign to R(4)R(4) so that RR is continuous?

Step 1: Write the mathematical expression

Simplify, then substitute t=4t = 4:

Key Takeaways

  • 1A discontinuity at x=ax = a means lim⁡x→af(x)=f(a)\lim_{x \to a} f(x) = f(a) fails
  • 2Removable (hole): the limit LL exists, but f(a)f(a) is missing or different; redefine f(a)=Lf(a) = L to fix it
  • 3Jump: finite one-sided limits that differ; jump size =∣lim⁡x→a+f(x)−lim⁡x→a−f(x)∣= \left| \lim_{x \to a^+} f(x) - \lim_{x \to a^-} f(x) \right|
  • 4Infinite: a one-sided limit is ±∞\pm\infty; the graph has a vertical asymptote x=ax = a
  • 5For rational functions, factor first: cancelled factors give holes, factors left in the denominator give asymptotes
  • 6The yy-coordinate of a hole comes from the simplified function, never from f(a)f(a)

Frequently Asked Questions

Because one point is enough to repair it. The limit LL already exists, so defining (or redefining) f(a)=Lf(a) = L makes lim⁡x→af(x)=f(a)\lim_{x \to a} f(x) = f(a) true, and the function becomes continuous at aa.
Because one point is enough to repair it. The limit LL already exists, so defining (or redefining) f(a)=Lf(a) = L makes lim⁡x→af(x)=f(a)\lim_{x \to a} f(x) = f(a) true, and the function becomes continuous at aa.
No. The left side heads to one height and the right side to another. Whatever value you give f(a)f(a), it can match at most one side, so the two-sided limit still does not exist.
No. The type depends only on the limits near aa. A jump stays a jump whether f(a)f(a) is the left value, the right value or undefined, and a hole stays removable whether f(a)f(a) is missing or wrong.
Yes, but they are rarer. For example sin⁡(1x)\sin\left(\frac{1}{x}\right) oscillates faster and faster near x=0x = 0, so it has no one-sided limits at all (an oscillating discontinuity). Jumps, infinite and oscillating discontinuities are all non-removable; many books call only the infinite and oscillating ones essential.

Glossary

Discontinuity
A point x=ax = a where lim⁡x→af(x)=f(a)\lim_{x \to a} f(x) = f(a) does not hold
Removable discontinuity (hole)
The limit exists and is finite, but f(a)f(a) is undefined or not equal to it; fixed by redefining one point
Jump discontinuity
Both one-sided limits are finite but not equal
Jump size
The distance between the two one-sided limits: ∣lim⁡x→a+f(x)−lim⁡x→a−f(x)∣\left| \lim_{x \to a^+} f(x) - \lim_{x \to a^-} f(x) \right|
Infinite discontinuity
At least one one-sided limit is ∞\infty or −∞-\infty; the graph has a vertical asymptote
Vertical asymptote
A vertical line x=ax = a that the graph approaches while its values grow without bound

Formula Card

Removable (hole)

lim⁡x→af(x)=L\lim_{x \to a} f(x) = L exists, f(a)≠Lf(a) \neq L or undefined

Fix it with $f(a) = L$

Jump

lim⁡x→a−f(x)≠lim⁡x→a+f(x)\lim_{x \to a^-} f(x) \neq \lim_{x \to a^+} f(x), both finite

Jump size $= \left| \lim_{x \to a^+} f(x) - \lim_{x \to a^-} f(x) \right|$

Infinite

lim⁡x→a±f(x)=±∞\lim_{x \to a^\pm} f(x) = \pm\infty

Vertical asymptote $x = a$

Rational function test

(x−a) p(x)(x−a) q(x)=p(x)q(x)\frac{(x - a)\,p(x)}{(x - a)\,q(x)} = \frac{p(x)}{q(x)}

A cancelled factor gives a hole at $\left(a, \frac{p(a)}{q(a)}\right)$ if $q(a) \neq 0$; a factor left in the denominator gives an asymptote

Today's challengeOne problem, three tries

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