Continuity | AP Calc AB Limits

What you’ll learn

The 3 conditions that define continuity at a point
How to classify discontinuities as removable, jump, or infinite
Intermediate value theorem (IVT)

Defining continuity with limits

A function $f (x)$ is continuous at a point $x = a$ if and only if it satisfies the following three conditions:

Definition of continuity

$f (a)$ is defined
$x \to a lim f (x)$ exists, meaning

$x \to a^{-} lim f (x) = x \to a^{+} lim f (x)$

$x \to a lim f (x) = f (a)$

In other words, the limit must exist and equal the function value. If any condition fails, the function is discontinuous at $x = a$ .

Questions similar to the following have previously appeared on the AP exam:

For what value of $k$ will the piecewise function $f (x)$ be continuous for all values of $x$ ?

$f (x) = ⎩ ⎨ ⎧ \frac{x ^{2} - 4 x + 3}{x - 1} k if x \neq = 1 if x = 1$

(spoiler)

For the rational function given, the only point where continuity could fail is at $x = 1$ , which is also the breakpoint.

From the piecewise function, $f (1) = k$ .

$f (x)$ follows the rational expression for all other values, $x \neq = 1$ , so the limit is

$x \to 1 lim \frac{x ^{2} - 4 x + 3}{x - 1} = x \to 1 lim \frac{( x - 1 ) ( x - 3 )}{( x - 1 )} = x \to 1 lim (x - 3) = - 2$

Since continuity requires $x \to 1 lim f (x) = f (1)$ ,

$k = - 2$

Classifying discontinuities

A point $x = a$ that does not satisfy conditions for continuity can be classified into one of three main categories:

1. Removable discontinuity (hole)

The limit exists but differs from the function value:

$x \to a lim f (x) \neq = f (a)$

$f (a)$ could be undefined or exist as a different value from the limit.

2. Jump discontinuity

The one-sided limits both exist but differ:

$x \to a^{-} lim f (x) \neq = x \to a^{+} lim f (x)$

$f (x)$ “jumps” from one value to another at $x = a$ .
Usually seen in piecewise functions at breakpoints.

3. Infinite (asymptotic) discontinuity

The limit is infinity:

$x \to a lim f (x) = \pm \infty$

Means $f (x)$ approaches a vertical asymptote at $x = a$ .

Shown below is a visual of each:

Example 1: Rational function

Classify any discontinuities of

$f (x) = \frac{x ^{2} - 4 x + 3}{x ^{2} - x - 6}$

1. Find where $f (x)$ is undefined

(spoiler)

The rational function is undefined when its denominator is $0$ :

$x^{2} - x - 6 = 0 (x + 2) (x - 3) = 0 x = - 2, 3$

2. Classify each discontinuity using limits

For $x = - 2$ :

(spoiler)

$x \to - 2 lim \frac{x ^{2} - 4 x + 3}{x ^{2} - x - 6} = x \to - 2 lim \frac{( x - 1 ) ( x - 3 )}{( x + 2 ) ( x - 3 )} = x \to - 2 lim \frac{( x - 1 )}{( x + 2 )}$

Direct substitution results in $\frac{- 3}{0}$ , which indicates a vertical asymptote. The one-sided limits confirm this:

$x \to - 2^{-} lim \frac{x - 1}{x + 2} x \to - 2^{+} lim \frac{x - 1}{x + 2} = \infty = - \infty$

Therefore, there is an infinite discontinuity at $x = - 2$ .

For $x = 3$ :

(spoiler)

$x \to 3 lim \frac{x ^{2} - 4 x + 3}{x ^{2} - x - 6} = x \to 3 lim \frac{( x - 1 )}{( x + 2 )} = \frac{2}{5}$

Since the limit exists but $f (3)$ is undefined, there is a removable discontinuity (hole) at $x = 3$ .

Example 2: Piecewise function

Let’s explore another example with a piecewise function:

Classify any discontinuities of $f (x)$ , where

$f (x) = ⎩ ⎨ ⎧ x + 2 1 - x + 4 if x < 1 if x = 1 if x > 1$

1. Identify potential discontinuities

(spoiler)

Since $x + 2$ and $- x + 4$ are both continuous, the breakpoint $x = 1$ is the only point where continuity could fail.

2. Analyze behavior using limits

(spoiler)

Left-hand limit:

$x \to 1^{-} lim (x + 2) = 3$

Right-hand limit:

$x \to 1^{+} lim (- x + 4) = 3$

Since the one-sided limits agree, $x \to 1 lim f (x) = 3$ .

However, because the piecewise function defines $f (1) = 1$ , then $x \to 1 lim f (x) \neq = f (1)$ .

Therefore, the discontinuity at $x = 1$ is removable.

Challenge problem

Find the values of $m$ and $b$ such that $f (x)$ continuous for all values of $x$ .

$f (x) = ⎩ ⎨ ⎧ m x^{2} + b 3 x - m x^{3} - b if x \leq - 2 if - 2 \leq x < 2 if x \geq 2$

Each piece is a polynomial that is continuous on its own interval. So the only points where continuity could fail are at the breakpoints of $x = - 2$ and $x = 2$ .

For $x = - 2$ :

(spoiler)

Continuity at $x = - 2$ requires

$f (- 2) = x \to - 2^{-} lim f (x) = x \to - 2^{+} lim f (x)$

The function value is

$f (- 2) = m (- 2)^{2} + b = 4 m + b$

The one-sided limits must be equal:

$x \to - 2^{-} lim f (x) x \to - 2^{-} lim (m x^{2} + b) 4 m + b 5 m + b = x \to - 2^{+} lim f (x) = x \to - 2^{+} lim (3 x - m) = - 6 - m = - 6$

For $x = 2$ :

(spoiler)

Continuity at $x = 2$ requires

$f (2) = x \to 2^{-} lim f (x) = x \to 2^{+} lim f (x)$

The function value is

$f (2) = (2)^{3} - b = 8 - b$

The one-sided limits must be equal:

$x \to 2^{-} lim f (x) x \to 2^{-} lim (3 x - m) 6 - m - m + b = x \to 2^{+} lim f (x) = x \to 2^{+} lim (x^{3} - b) = 8 - b = 2$

Now solve the resulting system of equations

(spoiler)

$5 m + b - m + b = - 6 = 2$

which gives $m = - \frac{4}{3}$ and $b = \frac{2}{3}$ .

Intermediate value theorem

One important consequence of continuity is the Intermediate Value Theorem (IVT).

Intermediate value theorem:

If $f (x)$ is continuous on the interval $[a, b]$ , and $L$ is any number between $f (a)$ and $f (b)$ , then there exists at least one number $c$ in $[a, b]$ such that $f (c) = L$ .

In plain language, a function that is continuous on $[a, b]$ can’t “skip over” values. It must take on every $y$ -value between $f (a)$ and $f (b)$ somewhere in that interval.

Examples

Let $f (x) = x^{2} - 4$ on the interval $[1, 3]$ . Use the IVT to show that there is a root in the interval.

(spoiler)

$f (x)$ is a polynomial, which is continuous for all $x$ .

A root means $f (x) = 0$ (which corresponds to crossing the $x$ -axis), so $L = 0$ .

Evaluate the endpoints:

$f (1) = 1^{2} - 4 = - 3$
$f (3) = 3^{2} - 4 = 5$

Because $- 3 < L < 5$ , the IVT guarantees at least one $c$ in $[1, 3]$ such that $f (c) = 0$ .

In this particular case, we can confirm by solving $f (c) = 0$ :

$c^{2} - 4 = 0 c = \pm 2$

Only $c = 2$ lies within the interval $[1, 3]$ .

AP tip:

For free-response questions that require justification using the Intermediate value theorem, you must explicitly state that

$f (x)$ is continuous on $[a, b]$ and
$f (a) < L < f (b)$ (or reversed, if $f (b) < f (a))$

to state that by the IVT, there must be a $c$ in $[a, b]$ such that $f (c) = L$ .

Let $f (x)$ be a continuous function defined on the interval $[1, 4]$ . The values of $f (x)$ at various points are given in the table:

$x$ $f (x)$

$1$ $2$

$2$ $5$

$3$ $3$

$4$ $7$

What is the minimum number of times $f (x)$ equals $4$ on the interval $[1, 4]$ ?

$x$	$f (x)$
$1$	$2$
$2$	$5$
$3$	$3$
$4$	$7$

(spoiler)

To find how many times $f (x) = 4$ , set the target value to $L = 4$ .

It’s often easiest to visualize the situation by sketching a possible graph through the points:

Because the function is continuous, the IVT can be applied to each subinterval. From the table:

On $[1, 2]$ , $f (1) = 2$ and $f (2) = 5$ , and $2 < L < 5$ .
On $[2, 3]$ , $f (2) = 5$ and $f (3) = 3$ , and $3 < L < 5$ .
On $[3, 4]$ , $f (3) = 3$ and $f (4) = 7$ , and $3 < L < 7$ .

So the minimum number of times $f (x) = 4$ is $3$ times.

What you’ll learn

The 3 conditions that define continuity at a point
How to classify discontinuities as removable, jump, or infinite
Intermediate value theorem (IVT)

Defining continuity with limits

A function $f (x)$ is continuous at a point $x = a$ if and only if it satisfies the following three conditions:

Definition of continuity

$f (a)$ is defined
$x \to a lim f (x)$ exists, meaning

$x \to a^{-} lim f (x) = x \to a^{+} lim f (x)$

$x \to a lim f (x) = f (a)$

In other words, the limit must exist and equal the function value. If any condition fails, the function is discontinuous at $x = a$ .

Questions similar to the following have previously appeared on the AP exam:

For what value of $k$ will the piecewise function $f (x)$ be continuous for all values of $x$ ?

$f (x) = ⎩ ⎨ ⎧ \frac{x ^{2} - 4 x + 3}{x - 1} k if x \neq = 1 if x = 1$

(spoiler)

For the rational function given, the only point where continuity could fail is at $x = 1$ , which is also the breakpoint.

From the piecewise function, $f (1) = k$ .

$f (x)$ follows the rational expression for all other values, $x \neq = 1$ , so the limit is

$x \to 1 lim \frac{x ^{2} - 4 x + 3}{x - 1} = x \to 1 lim \frac{( x - 1 ) ( x - 3 )}{( x - 1 )} = x \to 1 lim (x - 3) = - 2$

Since continuity requires $x \to 1 lim f (x) = f (1)$ ,

$k = - 2$

Classifying discontinuities

A point $x = a$ that does not satisfy conditions for continuity can be classified into one of three main categories:

1. Removable discontinuity (hole)

The limit exists but differs from the function value:

$x \to a lim f (x) \neq = f (a)$

$f (a)$ could be undefined or exist as a different value from the limit.

2. Jump discontinuity

The one-sided limits both exist but differ:

$x \to a^{-} lim f (x) \neq = x \to a^{+} lim f (x)$

$f (x)$ “jumps” from one value to another at $x = a$ .
Usually seen in piecewise functions at breakpoints.

3. Infinite (asymptotic) discontinuity

The limit is infinity:

$x \to a lim f (x) = \pm \infty$

Means $f (x)$ approaches a vertical asymptote at $x = a$ .

Shown below is a visual of each:

Example 1: Rational function

Classify any discontinuities of

$f (x) = \frac{x ^{2} - 4 x + 3}{x ^{2} - x - 6}$

1. Find where $f (x)$ is undefined

(spoiler)

The rational function is undefined when its denominator is $0$ :

$x^{2} - x - 6 = 0 (x + 2) (x - 3) = 0 x = - 2, 3$

2. Classify each discontinuity using limits

For $x = - 2$ :

(spoiler)

$x \to - 2 lim \frac{x ^{2} - 4 x + 3}{x ^{2} - x - 6} = x \to - 2 lim \frac{( x - 1 ) ( x - 3 )}{( x + 2 ) ( x - 3 )} = x \to - 2 lim \frac{( x - 1 )}{( x + 2 )}$

Direct substitution results in $\frac{- 3}{0}$ , which indicates a vertical asymptote. The one-sided limits confirm this:

$x \to - 2^{-} lim \frac{x - 1}{x + 2} x \to - 2^{+} lim \frac{x - 1}{x + 2} = \infty = - \infty$

Therefore, there is an infinite discontinuity at $x = - 2$ .

For $x = 3$ :

(spoiler)

$x \to 3 lim \frac{x ^{2} - 4 x + 3}{x ^{2} - x - 6} = x \to 3 lim \frac{( x - 1 )}{( x + 2 )} = \frac{2}{5}$

Since the limit exists but $f (3)$ is undefined, there is a removable discontinuity (hole) at $x = 3$ .

Example 2: Piecewise function

Let’s explore another example with a piecewise function:

Classify any discontinuities of $f (x)$ , where

$f (x) = ⎩ ⎨ ⎧ x + 2 1 - x + 4 if x < 1 if x = 1 if x > 1$

1. Identify potential discontinuities

(spoiler)

Since $x + 2$ and $- x + 4$ are both continuous, the breakpoint $x = 1$ is the only point where continuity could fail.

2. Analyze behavior using limits

(spoiler)

Left-hand limit:

$x \to 1^{-} lim (x + 2) = 3$

Right-hand limit:

$x \to 1^{+} lim (- x + 4) = 3$

Since the one-sided limits agree, $x \to 1 lim f (x) = 3$ .

However, because the piecewise function defines $f (1) = 1$ , then $x \to 1 lim f (x) \neq = f (1)$ .

Therefore, the discontinuity at $x = 1$ is removable.

Challenge problem

Find the values of $m$ and $b$ such that $f (x)$ continuous for all values of $x$ .

$f (x) = ⎩ ⎨ ⎧ m x^{2} + b 3 x - m x^{3} - b if x \leq - 2 if - 2 \leq x < 2 if x \geq 2$

Each piece is a polynomial that is continuous on its own interval. So the only points where continuity could fail are at the breakpoints of $x = - 2$ and $x = 2$ .

For $x = - 2$ :

(spoiler)

Continuity at $x = - 2$ requires

$f (- 2) = x \to - 2^{-} lim f (x) = x \to - 2^{+} lim f (x)$

The function value is

$f (- 2) = m (- 2)^{2} + b = 4 m + b$

The one-sided limits must be equal:

$x \to - 2^{-} lim f (x) x \to - 2^{-} lim (m x^{2} + b) 4 m + b 5 m + b = x \to - 2^{+} lim f (x) = x \to - 2^{+} lim (3 x - m) = - 6 - m = - 6$

For $x = 2$ :

(spoiler)

Continuity at $x = 2$ requires

$f (2) = x \to 2^{-} lim f (x) = x \to 2^{+} lim f (x)$

The function value is

$f (2) = (2)^{3} - b = 8 - b$

The one-sided limits must be equal:

$x \to 2^{-} lim f (x) x \to 2^{-} lim (3 x - m) 6 - m - m + b = x \to 2^{+} lim f (x) = x \to 2^{+} lim (x^{3} - b) = 8 - b = 2$

Now solve the resulting system of equations

(spoiler)

$5 m + b - m + b = - 6 = 2$

which gives $m = - \frac{4}{3}$ and $b = \frac{2}{3}$ .

Intermediate value theorem

One important consequence of continuity is the Intermediate Value Theorem (IVT).

Intermediate value theorem:

If $f (x)$ is continuous on the interval $[a, b]$ , and $L$ is any number between $f (a)$ and $f (b)$ , then there exists at least one number $c$ in $[a, b]$ such that $f (c) = L$ .

In plain language, a function that is continuous on $[a, b]$ can’t “skip over” values. It must take on every $y$ -value between $f (a)$ and $f (b)$ somewhere in that interval.

Examples

Let $f (x) = x^{2} - 4$ on the interval $[1, 3]$ . Use the IVT to show that there is a root in the interval.

(spoiler)

$f (x)$ is a polynomial, which is continuous for all $x$ .

A root means $f (x) = 0$ (which corresponds to crossing the $x$ -axis), so $L = 0$ .

Evaluate the endpoints:

$f (1) = 1^{2} - 4 = - 3$
$f (3) = 3^{2} - 4 = 5$

Because $- 3 < L < 5$ , the IVT guarantees at least one $c$ in $[1, 3]$ such that $f (c) = 0$ .

In this particular case, we can confirm by solving $f (c) = 0$ :

$c^{2} - 4 = 0 c = \pm 2$

Only $c = 2$ lies within the interval $[1, 3]$ .

AP tip:

For free-response questions that require justification using the Intermediate value theorem, you must explicitly state that

$f (x)$ is continuous on $[a, b]$ and
$f (a) < L < f (b)$ (or reversed, if $f (b) < f (a))$

to state that by the IVT, there must be a $c$ in $[a, b]$ such that $f (c) = L$ .

Let $f (x)$ be a continuous function defined on the interval $[1, 4]$ . The values of $f (x)$ at various points are given in the table:

$x$ $f (x)$

$1$ $2$

$2$ $5$

$3$ $3$

$4$ $7$

What is the minimum number of times $f (x)$ equals $4$ on the interval $[1, 4]$ ?

$x$	$f (x)$
$1$	$2$
$2$	$5$
$3$	$3$
$4$	$7$

(spoiler)

To find how many times $f (x) = 4$ , set the target value to $L = 4$ .

It’s often easiest to visualize the situation by sketching a possible graph through the points:

Because the function is continuous, the IVT can be applied to each subinterval. From the table:

On $[1, 2]$ , $f (1) = 2$ and $f (2) = 5$ , and $2 < L < 5$ .
On $[2, 3]$ , $f (2) = 5$ and $f (3) = 3$ , and $3 < L < 5$ .
On $[3, 4]$ , $f (3) = 3$ and $f (4) = 7$ , and $3 < L < 7$ .

So the minimum number of times $f (x) = 4$ is $3$ times.