AP Calculus AB and BC

Continuity and the Three Types of Discontinuity

Q: Is a removable discontinuity still a discontinuity if the limit exists?

Yes. The limit existing satisfies only the second condition of continuity. The function is still discontinuous at that point because f(c) is either undefined or not equal to the limit, so the graph has a hole. It becomes continuous only once you redefine f(c) to equal that limit.

Q: What is the difference between a jump and an infinite discontinuity?

Both have no overall limit, but for different reasons. At a jump the left and right one-sided limits are both finite, they just do not match, so the graph steps up or down. At an infinite discontinuity the function is unbounded, at least one one-sided limit runs off to +\infty or -\infty, which shows up as a vertical asymptote.

Q: Does the Intermediate Value Theorem tell you where the root is?

No. The IVT only guarantees that at least one such point exists somewhere in the open interval; it does not locate it and does not say how many there are. To find or estimate the actual value you need another method, such as solving the equation or a numerical technique.

A function is continuous at x = c when f(c) exists, the limit as x approaches c exists, and the two are equal. If any one fails you get a discontinuity: removable (a hole where the limit exists but misses f(c)), jump (left and right limits differ), or infinite (a vertical asymptote).

The three-part continuity test

A function $f$ is continuous at $x = c$ when three conditions all hold at once: $f(c)$ is defined, $\lim_{x \to c} f(x)$ exists, and those two values are equal (Topic 1.11). Check them in that order. If $f(c)$ is undefined there is nothing for the limit to match; if the limit does not exist there is no single value to match; and even when both exist, they can disagree. A discontinuity is just the name for the first condition that fails, and which condition fails, together with what the one-sided limits are doing, tells you the type.

f(c)\text{ is defined},\qquad \lim_{x \to c} f(x)\text{ exists},\qquad \lim_{x \to c} f(x) = f(c)

Graphically, continuity at $c$ means you can trace the graph straight through $x = c$ without lifting your pencil. A function is continuous on an interval when it is continuous at every point of that interval (Topic 1.12); at an endpoint you only need the appropriate one-sided limit. The standard function families, polynomial, rational, power, exponential, logarithmic, and trigonometric, are continuous at every point of their domains, so on the exam the only places worth checking are where a denominator equals zero or where a piecewise rule switches from one formula to another.

The three types of discontinuity

Every discontinuity the AB and BC exams ask you to classify is one of three kinds (Topic 1.10), and the quickest way to tell them apart is to ask what the two one-sided limits are doing.

Type	One-sided limits	Overall limit	Graph
Removable	Both equal the same finite number	Exists, but $\neq f(c)$ or $f(c)$ undefined	A single hole
Jump	Both finite but unequal	Does not exist	A break, the graph steps
Infinite	At least one is $+\infty$ or $-\infty$	Does not exist	A vertical asymptote

The removable discontinuity is the only one whose limit exists, and that single fact is worth memorizing. It is why this type, and only this type, can be repaired: the graph already agrees on the value it approaches from both sides, it just fails to actually take that value at $c$ . Jump and infinite discontinuities have no limit at all, so no single value can patch them.

In terms of the three-part test: a removable discontinuity fails only the last condition (or $f(c)$ is simply undefined), a jump fails the second condition because the one-sided limits disagree, and an infinite discontinuity also fails the second condition, with the function running off to infinity near $c$ .

How to spot which discontinuity you have

Recognizing which discontinuity you are looking at, before grinding through arithmetic, is the skill the exam is really testing. Two setups cover almost everything you will see.

Rational functions: factor the numerator and denominator completely, then examine each factor that makes the denominator zero. If that same factor also appears in the numerator and cancels completely, the discontinuity there is removable, a hole in the graph. If the factor remains in the denominator after canceling, the function grows without bound near that point and you have an infinite discontinuity, that is, a vertical asymptote. A single rational expression essentially never produces a jump.

Piecewise functions: the only points that can fail are the boundaries where the definition changes formula. At each boundary, compute the one-sided limit from the piece on each side. If the two one-sided limits agree with each other and with the function value there, $f$ is continuous. If both are finite but unequal, it is a jump. If either piece is unbounded, it is infinite. Jumps almost always come from piecewise definitions, so this is where to expect them.

Removing a discontinuity

When a limit exists at a discontinuity, you can remove it by redefining the function at that one point so it equals the limit (Topic 1.13). If $\lim_{x \to c} f(x) = L$ , then declaring $f(c) = L$ fills the hole and makes $f$ continuous at $c$ . This only works for removable discontinuities: jump and infinite discontinuities have no limit, so changing a single value cannot rescue them.

The same principle powers the common 'find the constant' problems. To make a piecewise function continuous at a boundary, set the expression from each side and the function value equal at that point, then solve for the unknown constant. Worked Example 2 runs this all the way through.

The Intermediate Value Theorem

The Intermediate Value Theorem is an existence theorem: it promises that a value is achieved somewhere without telling you where (Topic 1.16). If $f$ is continuous on the closed interval $[a, b]$ and $d$ is any number between $f(a)$ and $f(b)$ , then there is at least one $c$ in $(a, b)$ with $f(c) = d$ .

f\text{ continuous on }[a,b],\ \ d\text{ between }f(a)\text{ and }f(b)\ \implies\ f(c)=d\ \text{ for some }c\in(a,b)

Most IVT questions ask you to show that an equation has a solution. The routine is fixed: name a closed interval, state that $f$ is continuous on it, evaluate $f$ at both endpoints, and show the target value $d$ is caught between the two outputs, usually because $f$ changes sign so $d = 0$ sits between a negative and a positive value. Never skip the continuity step. A function with a jump can leap straight over $d$ without ever equaling it, so the theorem simply does not apply without continuity on the entire closed interval.

Common mistakes

A few traps show up again and again.

A hole is still a discontinuity. A limit existing does not make a function continuous; you also need $f(c)$ to exist and to equal that limit.
A vertical asymptote means the limit does not exist. Infinity is not a real number, so writing that a limit 'equals infinity' is only shorthand for the limit failing to exist.
The IVT gives you at least one $c$ , not exactly one, and it never tells you the value of $c$ .
The IVT needs continuity on the whole closed interval $[a, b]$ . Confirming the endpoints alone is not enough.

Worked examples

Worked example

Classify every discontinuity of a rational function

Find and classify all discontinuities of $f(x) = \dfrac{x^2 - 4}{x^2 - x - 2}$ .

Factor the numerator and denominator. The numerator is $x^2 - 4 = (x-2)(x+2)$ and the denominator is $x^2 - x - 2 = (x-2)(x+1)$ .
The denominator is zero at $x = 2$ and $x = -1$ , so those are the only places $f$ can be discontinuous. Everywhere else $f$ is a rational function on its domain, so it is continuous.
Cancel the common factor: $f(x) = \dfrac{(x-2)(x+2)}{(x-2)(x+1)} = \dfrac{x+2}{x+1}$ for $x \neq 2$ . The factor $(x-2)$ cancels completely, so the discontinuity at $x = 2$ is removable.
Find the limit at the hole: $\lim_{x \to 2} f(x) = \dfrac{2+2}{2+1} = \dfrac{4}{3}$ . There is a hole at $\left(2, \tfrac{4}{3}\right)$ .
At $x = -1$ the factor $(x+1)$ stays in the denominator. As $x \to -1$ the numerator approaches $1$ while the denominator approaches $0$ , so $f$ is unbounded. That is an infinite discontinuity, a vertical asymptote at $x = -1$ .

Removable discontinuity (hole) at x = 2 with limit 4/3, and an infinite discontinuity (vertical asymptote) at x = -1.

Worked example

Make a piecewise function continuous

Find the value of $c$ that makes $f$ continuous everywhere, where $f(x) = \begin{cases} cx + 1 & x \le 2 \\ x^2 - c & x > 2 \end{cases}$

Each piece is a polynomial, so the only place continuity can fail is the boundary $x = 2$ . Apply the three-part test there.
Function value: the first piece defines $f(2)$ , so $f(2) = c \cdot 2 + 1 = 2c + 1$ .
Left-hand limit: for $x \to 2^-$ use the first piece, so $\lim_{x \to 2^-} f(x) = 2c + 1$ .
Right-hand limit: for $x \to 2^+$ use the second piece, so $\lim_{x \to 2^+} f(x) = 2^2 - c = 4 - c$ .
For the overall limit to exist, the one-sided limits must be equal: $2c + 1 = 4 - c$ . Solve: $3c = 3$ , so $c = 1$ .
Check: with $c = 1$ , the left side gives $2(1)+1 = 3$ , the right side gives $4 - 1 = 3$ , and $f(2) = 3$ . All three agree, so $f$ is continuous. For any other value of $c$ the one-sided limits differ and $x = 2$ is a jump discontinuity.

c = 1. Any other value leaves a jump discontinuity at x = 2.

Worked example

Use the IVT to show a root exists

Show that $f(x) = x^3 + x - 1$ has at least one real root in the interval $[0, 1]$ .

Check the hypothesis first. $f$ is a polynomial, so it is continuous on the closed interval $[0, 1]$ , and the IVT applies.
Evaluate at the endpoints: $f(0) = 0^3 + 0 - 1 = -1$ and $f(1) = 1^3 + 1 - 1 = 1$ .
Identify the target value. A root means $f(c) = 0$ , so $d = 0$ . Since $f(0) = -1 < 0 < 1 = f(1)$ , the value $0$ lies between $f(0)$ and $f(1)$ .
Apply the IVT: because $f$ is continuous on $[0, 1]$ and $0$ is between $f(0)$ and $f(1)$ , there is at least one $c$ in $(0, 1)$ with $f(c) = 0$ . That $c$ is the root.

Because f is continuous on [0, 1] with f(0) = -1 and f(1) = 1, the IVT guarantees at least one c in (0, 1) where f(c) = 0.

Frequently asked questions

Is a removable discontinuity still a discontinuity if the limit exists?

Yes. The limit existing satisfies only the second condition of continuity. The function is still discontinuous at that point because $f(c)$ is either undefined or not equal to the limit, so the graph has a hole. It becomes continuous only once you redefine $f(c)$ to equal that limit.

What is the difference between a jump and an infinite discontinuity?

Both have no overall limit, but for different reasons. At a jump the left and right one-sided limits are both finite, they just do not match, so the graph steps up or down. At an infinite discontinuity the function is unbounded, at least one one-sided limit runs off to $+\infty$ or $-\infty$ , which shows up as a vertical asymptote.

Does the Intermediate Value Theorem tell you where the root is?

No. The IVT only guarantees that at least one such point exists somewhere in the open interval; it does not locate it and does not say how many there are. To find or estimate the actual value you need another method, such as solving the equation or a numerical technique.