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

In section 1.1, we looked at one-sided limits to determine if a limit exists overall. This idea can be extended to continuity - a function is defined to be continuous at a point $a$ if the limits from either side are the same and match the function’s value at that point. More formally, there are three conditions that must be met for a function to be continuous:

Definition of continuity at point $a$ :

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

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

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

If any of these conditions are not satisfied, the function is discontinuous at $x = a$ .

Example

This type of question often shows up on the AP exam:

What value of $k$ will make the piecewise function $f (x)$ continuous everywhere?

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

Solution

For $f (x)$ to be continuous, $x \to a lim f (x)$ and $f (a)$ must both exist and agree with each other.

Since $f (x)$ is a piecewise function, the breakpoint $x = 1$ is a potential “problem area” for the limit. This will be the $a$ we test for continuity at.

Condition 1: $f (a)$ is defined.

$f (1) = k$ as defined by the piecewise function.

Condition 2: $x \to 1 lim f (x)$ must exist.

$f (x)$ behaves like the rational function everywhere but $x = 1$ . So we must find

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

$= x \to 1 lim \frac{( x - 1 ) ( x - 3 )}{( x - 1 )}$

$= x \to 1 lim (x - 3)$

$= - 2$

Condition 3: $x \to 1 lim f (x)$ must match $f (1)$ for $f (x)$ to be continuous at $x = 1$ . So

$k = - 2$

This value for $k$ plugs in the hole created by the rational function.

Classifying discontinuities

The three main types of discontinuities are:

1. Removable discontinuity (hole)

When $f (x)$ passes condition 2 but fails condition 3.
i.e. $x \to a lim f (x)$ exists but is $\neq = f (a)$ .
- Removable discontinuities are often referred to as “holes.”

2. Jump discontinuity

When $f (x)$ fails condition 2
i.e. $x \to a^{-} lim f (x) \neq = x \to a^{+} lim f (x)$ as $f (x)$ “jumps” from one value to another at $x = a$ .
- Usually seen in piecewise functions at the breakpoint(s).

3. Infinite (asymptotic) discontinuity

When $x \to a lim f (x) = \pm \infty$ (does not exist). This means $f (x)$ approaches a vertical asymptote at $x = a$ .
- Often seen in rational functions where factoring and canceling still leaves a constant over a variable denominator i.e. $f (x) = \frac{1}{x - 4}$

Shown below is a visual of each:

If a question asks you to assess the continuity of a function and classify any discontinuities, follow these steps:

Check condition 1: Find potential discontinuities

Typically, discontinuities occur in:

Rational functions (when the denominator $= 0$ )
Logarithmic functions (since the argument must be $> 0$ )
Piecewise functions

If $f (a)$ is undefined, then $f (x)$ is discontinuous there. We just have to classify the discontinuity.

Even if $f (a)$ is defined (e.g. in a piecewise function) continuity is not guaranteed. Proceed to step 2.

Check condition 2: Classify discontinuity

Find the one-sided limits as $x \to a$ to classify the discontinuity.

If either one-sided limit is $\pm \infty$ , it’s an infinite discontinuity.
If the one-sided limits are finite but don’t match, it’s a jump discontinuity.
If they match and the limit exists, proceed to step 3.

Check condition 3: Limit $=$ f(a)

Match the results from conditions 1 and 2. If $f (a)$ from step 1 does not equal the limit from step 2, it’s a removable discontinuity.

If the values match, congrats! The function is continuous there.

Examples

1. Classify the discontinuities on $f (x) = \frac{x ^{2} - 4 x + 3}{x ^{2} - x - 6}$ , if any.

Solution

Answer: Infinite discontinuity at $x = 2$ , removable discontinuity at $x = 3.$

Condition 1: Potential discontinuities

This rational function is undefined when the denominator equals 0:

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

These are the points of discontinuity.

Condition 2: Classify

Check each point individually with conditions 2 and 3.

For $a = - 2$ :

Evaluate

$x \to - 2 lim \frac{x ^{2} - 4 x + 3}{x ^{2} - x - 6}$

Direct substitution results in the indeterminate form $\frac{0}{0}$ . Let’s try factoring both the numerator and the denominator and canceling:

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

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

Direct substitution yields $\frac{- 3}{0}$ , which suggests a vertical asymptote is involved. Checking the one-sided limits:

$Left: x \to - 2^{-} lim \frac{( x - 1 )}{( x + 2 )} = \infty$

$Right: x \to - 2^{+} lim \frac{( x - 1 )}{( x + 2 )} = - \infty$

Since the limits are infinite and the limit does not exist, there’s an infinite discontinuity at $x = 2$ .

For $a = 3$ :

Evaluate

$x \to 3 lim \frac{x ^{2} - 4 x + 3}{x ^{2} - x - 6}$

Using the same factoring technique, the limit becomes:

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

$= \frac{2}{5}$

The limit exists which passes condition 2.

Condition 3: Matching

Unfortunately $f (3)$ is still undefined so $x \to 3 lim f (x) \neq = f (3)$ and the point fails condition 3.

Therefore there is a removable discontinuity at $x = 3$ - a hole that, if filled in, would be at $y = \frac{2}{5}$ .

Let’s explore another example - a piecewise function:

2. Classify any discontinuities on $f (x)$ , the piecewise function defined by:

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

Solution

(spoiler)

Answer: Removable discontinuity at $x = 1$

Condition 1: $f (1) = 1$ as defined by the second piece. But since $x$ is approaching the breakpoint of the piecewise function in this limit, there is a potential discontinuity there.

Condition 2: Does $x \to 1^{-} lim f (x) = x \to 1^{+} lim f (x)?$

Left-hand limit:

As $x \to 1^{-}$ , $f (x)$ is defined by $(x + 2)$ for all $x < 1$ .

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

Right-hand limit:

As $x \to 1^{+}$ , $f (x)$ is defined by $(- x + 4)$ for all $x > 1$ .

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

Since the one-sided limits are the same finite number, $x \to 1 lim f (x) = 3$ and condition 2 has been satisfied.

Condition 3: Does $x \to 1 lim f (x) = f (1)$ ?

No, $x \to 1 lim f (x) = 3$ while $f (1) = 1$ . Therefore the discontinuity at $x = 1$ is removable.

Intermediate value theorem

One important property that results from 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 simpler terms, a continuous function will hit every value between $f (a)$ and $f (b)$ in the interval $[a, b]$ , including some $y$ -value $L$ at some $x$ -value $c$ .

Examples

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

Solution

$f (x)$ is a polynomial so continuous everywhere.

A root, or solution, means that $f (x)$ crosses the $x$ -axis (where $L = 0$ ).

Check the values at the endpoints:

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

Because $- 3 < L < 5$ , the IVT guarantees some $c$ in $[1, 3]$ where $f (c) = 0$ , which means $c$ is a root.

To find $c$ :

$f (c) = 0$ means that

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

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

AP tip:

If a free-response question requires the use of the IVT, make sure to explicitly state the conditions met - that $f (x)$ is continuous and the target value $L$ is between $f (a)$ and $f (b)$ - to conclude the existence of $c$ in $[a, b]$ where $f (c) = L$ .

2. 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 $f (x) = 4$ on the interval $[1, 4]$ ?

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

Solution

(spoiler)

Answer: $3 times$

While we could apply the IVT on each interval $[1, 2]$ , $[2, 3]$ , and $[3, 4]$ , it would be easier to just draw a graph for a general idea of the situation:

The four points are what’s important - the function can look much crazier and quickly bounce back and forth between multiple values. However,

f (x)

has to pass the horizontal line

L = 4

at the very least

3

times, in order to go from a point below the line to a point above it and vice versa.

Challenge problem

1. For what values of $m$ and $b$ will $f (x)$ be continuous over all real numbers?

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

Solution

(spoiler)

Answer: $m = - \frac{4}{3} and b = \frac{2}{3}$

All 3 pieces are polynomials and continuous over their individual intervals. The potential discontinuities are at the breakpoints.

Breakpoint $x = - 2$ :

For $f (x)$ to be continuous at $x = - 2$ ,

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

First,

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

Next, the one-sided limits:

Left: $x \to - 2^{-} lim f (x)$

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

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

$= 3 (- 2) - m = - 6 - m$

The one-sided limits must match, so setting them equal to each other and rearranging,

$5 m + b = - 6$

However there are two unknown variables, so we need another equation that relates $m$ and $b$ .

Breakpoint $x = 2$ :

This time, we need

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

for $f (x)$ to be continuous at $x = 2$ .

First,

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

Next, the one-sided limits:

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

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

They must match, so setting them equal and rearranging,

$m - b = - 2$

Solving the system of equations

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

results in $m = - \frac{4}{3}$ and $b = \frac{2}{3}$ .

Key points

A function is continuous at $x = a$ if, in short:
$f (a) = x \to a lim f (x)$ (both exist and match).
If not, there are 3 main types of discontinuities:

Type of discontinuity	Occurs when $f (x)$ …
Removable (hole)	Passes cond. 2 but fails cond. 3: Limit $\neq = f (a)$
Jump	Fails cond. 2: Limit DNE
Infinite/asymptotic	Fails cond. 2: Limit $= \pm \infty$

The Intermediate value theorem is a guarantee that a continuous function will take all intermediate values between $f (a)$ and $f (b)$ , making it useful for proving the existence of a root or some specified value, even if the solution can’t be found easily.

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

Definition of continuity at point $a$ :

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

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

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

If any of these conditions are not satisfied, the function is discontinuous at $x = a$ .

Example

This type of question often shows up on the AP exam:

What value of $k$ will make the piecewise function $f (x)$ continuous everywhere?

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

Solution

For $f (x)$ to be continuous, $x \to a lim f (x)$ and $f (a)$ must both exist and agree with each other.

Since $f (x)$ is a piecewise function, the breakpoint $x = 1$ is a potential “problem area” for the limit. This will be the $a$ we test for continuity at.

Condition 1: $f (a)$ is defined.

$f (1) = k$ as defined by the piecewise function.

Condition 2: $x \to 1 lim f (x)$ must exist.

$f (x)$ behaves like the rational function everywhere but $x = 1$ . So we must find

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

$= x \to 1 lim \frac{( x - 1 ) ( x - 3 )}{( x - 1 )}$

$= x \to 1 lim (x - 3)$

$= - 2$

Condition 3: $x \to 1 lim f (x)$ must match $f (1)$ for $f (x)$ to be continuous at $x = 1$ . So

$k = - 2$

This value for $k$ plugs in the hole created by the rational function.

Classifying discontinuities

The three main types of discontinuities are:

1. Removable discontinuity (hole)

When $f (x)$ passes condition 2 but fails condition 3.
i.e. $x \to a lim f (x)$ exists but is $\neq = f (a)$ .
- Removable discontinuities are often referred to as “holes.”

2. Jump discontinuity

When $f (x)$ fails condition 2
i.e. $x \to a^{-} lim f (x) \neq = x \to a^{+} lim f (x)$ as $f (x)$ “jumps” from one value to another at $x = a$ .
- Usually seen in piecewise functions at the breakpoint(s).

3. Infinite (asymptotic) discontinuity

When $x \to a lim f (x) = \pm \infty$ (does not exist). This means $f (x)$ approaches a vertical asymptote at $x = a$ .
- Often seen in rational functions where factoring and canceling still leaves a constant over a variable denominator i.e. $f (x) = \frac{1}{x - 4}$

Shown below is a visual of each:

If a question asks you to assess the continuity of a function and classify any discontinuities, follow these steps:

Check condition 1: Find potential discontinuities

Typically, discontinuities occur in:

Rational functions (when the denominator $= 0$ )
Logarithmic functions (since the argument must be $> 0$ )
Piecewise functions

If $f (a)$ is undefined, then $f (x)$ is discontinuous there. We just have to classify the discontinuity.

Even if $f (a)$ is defined (e.g. in a piecewise function) continuity is not guaranteed. Proceed to step 2.

Check condition 2: Classify discontinuity

Find the one-sided limits as $x \to a$ to classify the discontinuity.

If either one-sided limit is $\pm \infty$ , it’s an infinite discontinuity.
If the one-sided limits are finite but don’t match, it’s a jump discontinuity.
If they match and the limit exists, proceed to step 3.

Check condition 3: Limit $=$ f(a)

Match the results from conditions 1 and 2. If $f (a)$ from step 1 does not equal the limit from step 2, it’s a removable discontinuity.

If the values match, congrats! The function is continuous there.

Examples

1. Classify the discontinuities on $f (x) = \frac{x ^{2} - 4 x + 3}{x ^{2} - x - 6}$ , if any.

Solution

Answer: Infinite discontinuity at $x = 2$ , removable discontinuity at $x = 3.$

Condition 1: Potential discontinuities

This rational function is undefined when the denominator equals 0:

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

These are the points of discontinuity.

Condition 2: Classify

Check each point individually with conditions 2 and 3.

For $a = - 2$ :

Evaluate

$x \to - 2 lim \frac{x ^{2} - 4 x + 3}{x ^{2} - x - 6}$

Direct substitution results in the indeterminate form $\frac{0}{0}$ . Let’s try factoring both the numerator and the denominator and canceling:

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

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

Direct substitution yields $\frac{- 3}{0}$ , which suggests a vertical asymptote is involved. Checking the one-sided limits:

$Left: x \to - 2^{-} lim \frac{( x - 1 )}{( x + 2 )} = \infty$

$Right: x \to - 2^{+} lim \frac{( x - 1 )}{( x + 2 )} = - \infty$

Since the limits are infinite and the limit does not exist, there’s an infinite discontinuity at $x = 2$ .

For $a = 3$ :

Evaluate

$x \to 3 lim \frac{x ^{2} - 4 x + 3}{x ^{2} - x - 6}$

Using the same factoring technique, the limit becomes:

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

$= \frac{2}{5}$

The limit exists which passes condition 2.

Condition 3: Matching

Unfortunately $f (3)$ is still undefined so $x \to 3 lim f (x) \neq = f (3)$ and the point fails condition 3.

Therefore there is a removable discontinuity at $x = 3$ - a hole that, if filled in, would be at $y = \frac{2}{5}$ .

Let’s explore another example - a piecewise function:

2. Classify any discontinuities on $f (x)$ , the piecewise function defined by:

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

Solution

(spoiler)

Answer: Removable discontinuity at $x = 1$

Condition 1: $f (1) = 1$ as defined by the second piece. But since $x$ is approaching the breakpoint of the piecewise function in this limit, there is a potential discontinuity there.

Condition 2: Does $x \to 1^{-} lim f (x) = x \to 1^{+} lim f (x)?$

Left-hand limit:

As $x \to 1^{-}$ , $f (x)$ is defined by $(x + 2)$ for all $x < 1$ .

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

Right-hand limit:

As $x \to 1^{+}$ , $f (x)$ is defined by $(- x + 4)$ for all $x > 1$ .

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

Since the one-sided limits are the same finite number, $x \to 1 lim f (x) = 3$ and condition 2 has been satisfied.

Condition 3: Does $x \to 1 lim f (x) = f (1)$ ?

No, $x \to 1 lim f (x) = 3$ while $f (1) = 1$ . Therefore the discontinuity at $x = 1$ is removable.

Intermediate value theorem

One important property that results from 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 simpler terms, a continuous function will hit every value between $f (a)$ and $f (b)$ in the interval $[a, b]$ , including some $y$ -value $L$ at some $x$ -value $c$ .

Examples

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

Solution

$f (x)$ is a polynomial so continuous everywhere.

A root, or solution, means that $f (x)$ crosses the $x$ -axis (where $L = 0$ ).

Check the values at the endpoints:

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

Because $- 3 < L < 5$ , the IVT guarantees some $c$ in $[1, 3]$ where $f (c) = 0$ , which means $c$ is a root.

To find $c$ :

$f (c) = 0$ means that

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

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

AP tip:

2. 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 $f (x) = 4$ on the interval $[1, 4]$ ?

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

Solution

(spoiler)

Answer: $3 times$

While we could apply the IVT on each interval $[1, 2]$ , $[2, 3]$ , and $[3, 4]$ , it would be easier to just draw a graph for a general idea of the situation:

The four points are what’s important - the function can look much crazier and quickly bounce back and forth between multiple values. However,

f (x)

has to pass the horizontal line

L = 4

at the very least

3

times, in order to go from a point below the line to a point above it and vice versa.

Challenge problem

1. For what values of $m$ and $b$ will $f (x)$ be continuous over all real numbers?

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

Solution

(spoiler)

Answer: $m = - \frac{4}{3} and b = \frac{2}{3}$

All 3 pieces are polynomials and continuous over their individual intervals. The potential discontinuities are at the breakpoints.

Breakpoint $x = - 2$ :

For $f (x)$ to be continuous at $x = - 2$ ,

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

First,

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

Next, the one-sided limits:

Left: $x \to - 2^{-} lim f (x)$

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

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

$= 3 (- 2) - m = - 6 - m$

The one-sided limits must match, so setting them equal to each other and rearranging,

$5 m + b = - 6$

However there are two unknown variables, so we need another equation that relates $m$ and $b$ .

Breakpoint $x = 2$ :

This time, we need

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

for $f (x)$ to be continuous at $x = 2$ .

First,

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

Next, the one-sided limits:

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

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

They must match, so setting them equal and rearranging,

$m - b = - 2$

Solving the system of equations

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

results in $m = - \frac{4}{3}$ and $b = \frac{2}{3}$ .

Key points

A function is continuous at $x = a$ if, in short:
$f (a) = x \to a lim f (x)$ (both exist and match).
If not, there are 3 main types of discontinuities:

Type of discontinuity	Occurs when $f (x)$ …
Removable (hole)	Passes cond. 2 but fails cond. 3: Limit $\neq = f (a)$
Jump	Fails cond. 2: Limit DNE
Infinite/asymptotic	Fails cond. 2: Limit $= \pm \infty$

The Intermediate value theorem is a guarantee that a continuous function will take all intermediate values between $f (a)$ and $f (b)$ , making it useful for proving the existence of a root or some specified value, even if the solution can’t be found easily.

Continuity

What you’ll learn:

Defining continuity with limits

Example

Solution

Classifying discontinuities

Examples

Solution

Solution

Intermediate value theorem

Examples

Solution

Solution

Challenge problem

Solution

Continuity

What you’ll learn:

Defining continuity with limits

Example

Solution

Classifying discontinuities

Examples

Solution

Solution

Intermediate value theorem

Examples

Solution

Solution

Challenge problem

Solution