Functions & Continuity

Q: What causes a removable discontinuity?

A common factor that cancels from the numerator and denominator. The point is missing from the domain, leaving a hole even though the simplified function is otherwise fine there.

Q: How is a jump different from an infinite discontinuity?

A jump is a finite step between two pieces; an infinite discontinuity shoots off toward $\pm\infty$ at a vertical asymptote.

Q: Are polynomials ever discontinuous?

No. Every polynomial is continuous everywhere — no denominators to hit zero and no piece boundaries.

A function is continuous if you can draw its graph without lifting your pencil. When you have to lift the pencil, there is a discontinuity — a break in the graph.

Discontinuities come in three flavors: removable holes, jumps, and infinite breaks. Naming them correctly is the goal here, and each has a telltale cause in the equation.

The three kinds of discontinuity

A removable discontinuity is a single missing point — a hole — usually from a factor that cancels. A jump discontinuity is a sudden step, common in piecewise functions.

An infinite discontinuity happens at a vertical asymptote, where the graph shoots up or down without bound. The function value blows up near that x.

The graph below has a jump at $x = 1$ : the left piece ends at an open dot and the right piece restarts lower down, so you must lift your pencil to keep drawing.

Spotting them from an equation

A denominator that equals zero but does not cancel produces a vertical asymptote and an infinite discontinuity. A factor that cancels top and bottom leaves a hole.

Piecewise definitions that do not meet up at their boundary create a jump. Check the two pieces at the boundary x-value to see whether they agree.

Worked examples

Example 1: naming an asymptote's break

Which type of discontinuity does a vertical asymptote create?

Near the asymptote the graph is unbounded

|f(x)| \to \infty

That is the infinite type

\text{infinite discontinuity}

Answer: An infinite discontinuity

Example 2: a hole

Why does $f(x) = \dfrac{(x-2)(x+1)}{x-2}$ have a hole at $x = 2$ ?

The factor cancels

\dfrac{(x-2)(x+1)}{x-2} = x + 1

But x = 2 was never allowed

x \neq 2

So there is a single missing point

\text{hole at } x = 2

Answer: A removable discontinuity (hole) at $x = 2$

Example 3: a jump from a piecewise rule

For $f(x) = x + 1$ when $x < 1$ and $f(x) = x - 2$ when $x \geq 1$ , what happens at $x = 1$ ?

Check where the first piece ends

1 + 1 = 2

Check where the second piece starts

1 - 2 = -1

The pieces do not meet, so the graph steps

\text{jump discontinuity}

Answer: A jump discontinuity at $x = 1$

Try one yourself

Common questions

What causes a removable discontinuity?

A common factor that cancels from the numerator and denominator. The point is missing from the domain, leaving a hole even though the simplified function is otherwise fine there.

How is a jump different from an infinite discontinuity?

A jump is a finite step between two pieces; an infinite discontinuity shoots off toward $\pm\infty$ at a vertical asymptote.

Are polynomials ever discontinuous?

No. Every polynomial is continuous everywhere — no denominators to hit zero and no piece boundaries.

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Functions & Continuity

The three kinds of discontinuity

Spotting them from an equation

Worked examples

Example 1: naming an asymptote's break

Example 2: a hole

Example 3: a jump from a piecewise rule

Try one yourself

Common questions

Keep learning

Key Features of Functions

Domain & Range from Graphs and Equations

Graphing Rational Functions