1.2.2 Composite functions

Achievable AP Calculus AB

1. Limits

1.2. Analytical limits

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Composite functions

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What you’ll learn

Composite functions: Find limits of composite functions algebraically.
Graphical composition: Determine composite limits using given graphs.

A composite function is a function inside another function, usually written as $f (g (x))$ or occasionally as $(f \circ g) (x)$ .

Limit of a composite function:

$x \to a lim g (x) = L$

and $f (x)$ is continuous at $x = L$ , then

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

If $f$ is not continuous at $L$ , one-sided limits must be used to analyze the behavior.

So to find the limit of a composite function,

Find the limit $L$ of the inner function $g (x)$ .
Plug that result into the outer function $f$ , provided $f$ is continuous at $L$ .

Example 1: Continuous functions

If $f (x) = 3 x + 1$ and $g (x) = 2 x^{2}$ , evaluate

a) $x \to 0 lim f (g (x))$

b) $x \to 0 lim g (f (x))$

Solutions

a) $x \to 0 lim f (g (x))$

(spoiler)

First, evaluate the limit of the inner function $g (x)$ to find $L$ :

$x \to 0 lim g (x) = x \to 0 lim 2 x^{2} = 0$

Now evaluate the outer function $f$ at $L = 0$ :

$f (0) = 3 (0) + 1 = 1$

b) $x \to 0 lim g (f (x))$

(spoiler)

First, evaluate the limit of the inner function $f (x)$ to find $L$ :

$x \to 0 lim f (x) = x \to 0 lim (3 x + 1) = 1$

Now evaluate the outer function $g$ at $L = 1$ :

$g (1) = 2 (1)^{2} = 2$

Example 2: Discontinuities

The composite-function limit law works when the outer function $f (x)$ is continuous at the value approached by the inner function $g (x)$ , allowing direct substitution into the outer function.

But if $x \to a lim g (x)$ doesn’t exist or if $f (x)$ is not continuous at $x = L$ , use one-sided limits to analyze the behaviors of each function further.

If $f (x) = \frac{1}{x}$ and $g (x) = x^{2} - 9$ , evaluate:

$x \to 3 lim f (g (x))$

Solution

(spoiler)

First, evaluate the limit of the inner function to find $L$ :

$x \to 3 lim g (x) = x \to 3 lim (x^{2} - 9) = 0$

Because the outer function $f (x) = \frac{1}{x}$ is not continuous at the result of $L = 0$ , we must analyze how the inner function $g (x)$ approaches $0$ using one-sided limits to determine how the outer $f$ behaves.

You can follow along with the explanation by graphing the two functions in Desmos. Enter the following lines:

$g (x) f (x) = x^{2} - 9 = \frac{1}{x}$

Now, observe how $g (x)$ behaves as $x \to 3$ from either side.

1. From the left:

As $x \to 3^{-}$ , the graph of $g$ approaches $0$ from below. For example,
- $g (2.9) = - 0.59$
- $g (2.99) = - 0.0599$
Inputting these negative values into $f (x) = \frac{1}{x}$ gives values that decrease without bound toward $- \infty$ . So:

$x \to 3^{-} lim f (g (x)) = - \infty$

2. From the right:

As $x \to 3^{+}$ , the graph of $g (x)$ approaches $0$ from above (e.g. $g (3.1) = 0.61$ ).
Inputting these positive values into $f (x)$ gives values that increase without bound toward $\infty$ . So:

$x \to 3^{+} lim f (g (x)) = \infty$

Because the two one-sided limits are not equal,

$x \to 3 lim f (g (x)) does not exist.$

You can visualize this by typing $f (g (x))$ directly into Desmos and observing the behavior of the composite function to verify the limits found algebraically.

Evaluating composite limits graphically

Problems similar to the multi-part question below have appeared on past AP exams, including some of the trickier but less common cases.

In the graphs shown below, $g (x)$ has a vertical asymptote at $x = 4$ . The function $f (x)$ is defined only for select values within the interval $(- 5, 8]$ .

Composite function limit

Evaluate:

a) $x \to 4 lim f (g (x))$

b) $x \to 1 lim g (f (x))$

c) $x \to - 2 lim f (f (x))$

d) $x \to - 1 lim g (g (x))$

Answers

(spoiler)

a) Does not exist
b) $\infty$
c) $2$
d) Does not exist

Solutions

a) $x \to 4 lim f (g (x))$

(spoiler)

As $x$ approaches $4$ , the inner function $g (x)$ approaches the vertical asymptote, increasing without bound to $\infty$ on both sides.

However, the outer function $f (x)$ is defined only for inputs in the interval $(- 5, 8]$ . For values of $x$ close to $4$ , $g (x)$ lies outside of the domain of $f$ .

For example, if $g (3.99) = 1000$ then $f (g (3.99)) = f (1000)$ , which is not defined.

Since the composite function $f (g (x))$ is not defined for $x$ -values close to $4$ , the limit does not exist.

b) $x \to 1 lim g (f (x))$

(spoiler)

First, evaluate the inner limit to find $L$ :

$x \to 1 lim f (x) = 4$

At $x = 4$ , the outer function $g$ has a vertical asymptote, so it would be reasonable to conclude the composite limit is $\infty$ . But more specifically, the behavior of $g$ can be described using one-sided limits.

1. From the left:

As $x \to 1^{-}$ , the graph shows $f (x)$ approaching $4$ from below. For example, we might have values such as

$f (0.99) = 3.99$
$f (0.999) = 3.999$ , etc.

When these values are inputted into $g$ , such as $g (3.99)$ , $g (3.999)$ , etc. the output grows without bound, and

$x \to 4^{-} lim g (x) = \infty$

2. From the right:

Similarly, as $x \to 1^{+}$ , $f$ also approaches $4$ from below, and $x \to 4^{+} lim g (x) = \infty$ .

Therefore, the composite limit is

$x \to 1 lim g (f (x)) = \infty$

c) $x \to - 2 lim f (f (x))$

(spoiler)

First, evaluate the inner limit:

$x \to - 2 lim f (x) = 3$

Since $f$ is not continuous at $x = 3$ , simply substituting $f (3) = - 3$ does not determine the limit. Instead, we need to analyze how $f (x)$ approaches $3$ .

From the graph, as $x \to - 2$ from either side, $f (x)$ approaches $3$ from above. For example, we might have values such as

$f (- 1.99) = 3.01$
$f (- 2.01) = 3.001$

These values are all slightly greater than $3$ , meaning the outer function is evaluated at inputs to the right of $3$ , and the composite function approaches the right-hand limit of $f$ :

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

Therefore, the composite limit is

$x \to - 2 lim f (f (x)) = 2$

d) $x \to - 1 lim g (g (x))$

(spoiler)

First, evaluate the inner limit:

$x \to - 1 lim g (x) = - 4$

Since $g$ is not continuous at $x = - 4$ , we must analyze how $g$ approaches $- 4$ as $x$ approaches $- 1$ , using one-sided limits.

1. From the left:

As $x \to - 1^{-}$ , the inner function $g (x)$ approaches $- 4$ from below. For example, we may have values such as

$g (- 1.01) = - 4.01$
$g (- 1.001) = - 4.001$

These values are all slightly less than $- 4$ , meaning the outer $g$ is evaluated at inputs to the left of $- 4$ . From the graph, the left-hand limit is

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

2. From the right:

As $x \to - 1^{+}$ , $g (x)$ approaches $- 4$ from above. For example, we might have $g (- 0.99) = - 3.99$ .

These values are all slightly greater $- 4$ , meaning the outer $g$ is evaluated at inputs to the right $- 4$ . From the graph, the right-hand limit is

$x \to - 4^{+} lim g (x) = - 2$

Because the one-sided limits of $g$ as $x$ approaches $- 4$ differ, the one-sided limits of $g (g (x))$ as $x$ approaches $- 1$ also differ, so

$x \to - 1 lim g (g (x)) does not exist.$

Composite functions

Form: $f (g (x))$ or $(f \circ g) (x)$ — a function inside another function
Limit rule: $lim_{x \to a} f (g (x)) = f (lim_{x \to a} g (x)) = f (L)$
- Requires $f (x)$ to be continuous at $x = L$

Steps to evaluate composite limits

Step 1: Find the limit of the inner function
Step 2: Plug that result into the outer function

When the standard rule fails

If $lim_{x \to a} g (x)$ puts values outside the domain of $f$ → limit DNE
If $f$ is discontinuous at $L$ , substituting $f (L)$ is not valid — must analyze one-sided behavior instead

One-sided analysis for tricky cases

Determine whether the inner function approaches $L$ from above or below
Use the corresponding one-sided limit of the outer function
If left- and right-hand composite limits differ → overall limit DNE

Composite functions

What you’ll learn

Composite functions: Find limits of composite functions algebraically.
Graphical composition: Determine composite limits using given graphs.

A composite function is a function inside another function, usually written as $f (g (x))$ or occasionally as $(f \circ g) (x)$ .

Limit of a composite function:

$x \to a lim g (x) = L$

and $f (x)$ is continuous at $x = L$ , then

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

If $f$ is not continuous at $L$ , one-sided limits must be used to analyze the behavior.

So to find the limit of a composite function,

Find the limit $L$ of the inner function $g (x)$ .
Plug that result into the outer function $f$ , provided $f$ is continuous at $L$ .

Example 1: Continuous functions

If $f (x) = 3 x + 1$ and $g (x) = 2 x^{2}$ , evaluate

a) $x \to 0 lim f (g (x))$

b) $x \to 0 lim g (f (x))$

Solutions

a) $x \to 0 lim f (g (x))$

(spoiler)

First, evaluate the limit of the inner function $g (x)$ to find $L$ :

$x \to 0 lim g (x) = x \to 0 lim 2 x^{2} = 0$

Now evaluate the outer function $f$ at $L = 0$ :

$f (0) = 3 (0) + 1 = 1$

b) $x \to 0 lim g (f (x))$

(spoiler)

First, evaluate the limit of the inner function $f (x)$ to find $L$ :

$x \to 0 lim f (x) = x \to 0 lim (3 x + 1) = 1$

Now evaluate the outer function $g$ at $L = 1$ :

$g (1) = 2 (1)^{2} = 2$

Example 2: Discontinuities

The composite-function limit law works when the outer function $f (x)$ is continuous at the value approached by the inner function $g (x)$ , allowing direct substitution into the outer function.

But if $x \to a lim g (x)$ doesn’t exist or if $f (x)$ is not continuous at $x = L$ , use one-sided limits to analyze the behaviors of each function further.

If $f (x) = \frac{1}{x}$ and $g (x) = x^{2} - 9$ , evaluate:

$x \to 3 lim f (g (x))$

Solution

(spoiler)

First, evaluate the limit of the inner function to find $L$ :

$x \to 3 lim g (x) = x \to 3 lim (x^{2} - 9) = 0$

You can follow along with the explanation by graphing the two functions in Desmos. Enter the following lines:

$g (x) f (x) = x^{2} - 9 = \frac{1}{x}$

Now, observe how $g (x)$ behaves as $x \to 3$ from either side.

1. From the left:

As $x \to 3^{-}$ , the graph of $g$ approaches $0$ from below. For example,
- $g (2.9) = - 0.59$
- $g (2.99) = - 0.0599$
Inputting these negative values into $f (x) = \frac{1}{x}$ gives values that decrease without bound toward $- \infty$ . So:

$x \to 3^{-} lim f (g (x)) = - \infty$

2. From the right:

As $x \to 3^{+}$ , the graph of $g (x)$ approaches $0$ from above (e.g. $g (3.1) = 0.61$ ).
Inputting these positive values into $f (x)$ gives values that increase without bound toward $\infty$ . So:

$x \to 3^{+} lim f (g (x)) = \infty$

Because the two one-sided limits are not equal,

$x \to 3 lim f (g (x)) does not exist.$

You can visualize this by typing $f (g (x))$ directly into Desmos and observing the behavior of the composite function to verify the limits found algebraically.

Evaluating composite limits graphically

Problems similar to the multi-part question below have appeared on past AP exams, including some of the trickier but less common cases.

In the graphs shown below, $g (x)$ has a vertical asymptote at $x = 4$ . The function $f (x)$ is defined only for select values within the interval $(- 5, 8]$ .

Evaluate:

a) $x \to 4 lim f (g (x))$

b) $x \to 1 lim g (f (x))$

c) $x \to - 2 lim f (f (x))$

d) $x \to - 1 lim g (g (x))$

Answers

(spoiler)

a) Does not exist
b) $\infty$
c) $2$
d) Does not exist

Solutions

a) $x \to 4 lim f (g (x))$

(spoiler)

As $x$ approaches $4$ , the inner function $g (x)$ approaches the vertical asymptote, increasing without bound to $\infty$ on both sides.

However, the outer function $f (x)$ is defined only for inputs in the interval $(- 5, 8]$ . For values of $x$ close to $4$ , $g (x)$ lies outside of the domain of $f$ .

For example, if $g (3.99) = 1000$ then $f (g (3.99)) = f (1000)$ , which is not defined.

Since the composite function $f (g (x))$ is not defined for $x$ -values close to $4$ , the limit does not exist.

b) $x \to 1 lim g (f (x))$

(spoiler)

First, evaluate the inner limit to find $L$ :

$x \to 1 lim f (x) = 4$

1. From the left:

As $x \to 1^{-}$ , the graph shows $f (x)$ approaching $4$ from below. For example, we might have values such as

$f (0.99) = 3.99$
$f (0.999) = 3.999$ , etc.

When these values are inputted into $g$ , such as $g (3.99)$ , $g (3.999)$ , etc. the output grows without bound, and

$x \to 4^{-} lim g (x) = \infty$

2. From the right:

Similarly, as $x \to 1^{+}$ , $f$ also approaches $4$ from below, and $x \to 4^{+} lim g (x) = \infty$ .

Therefore, the composite limit is

$x \to 1 lim g (f (x)) = \infty$

c) $x \to - 2 lim f (f (x))$

(spoiler)

First, evaluate the inner limit:

$x \to - 2 lim f (x) = 3$

Since $f$ is not continuous at $x = 3$ , simply substituting $f (3) = - 3$ does not determine the limit. Instead, we need to analyze how $f (x)$ approaches $3$ .

From the graph, as $x \to - 2$ from either side, $f (x)$ approaches $3$ from above. For example, we might have values such as

$f (- 1.99) = 3.01$
$f (- 2.01) = 3.001$

These values are all slightly greater than $3$ , meaning the outer function is evaluated at inputs to the right of $3$ , and the composite function approaches the right-hand limit of $f$ :

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

Therefore, the composite limit is

$x \to - 2 lim f (f (x)) = 2$

d) $x \to - 1 lim g (g (x))$

(spoiler)

First, evaluate the inner limit:

$x \to - 1 lim g (x) = - 4$

Since $g$ is not continuous at $x = - 4$ , we must analyze how $g$ approaches $- 4$ as $x$ approaches $- 1$ , using one-sided limits.

1. From the left:

As $x \to - 1^{-}$ , the inner function $g (x)$ approaches $- 4$ from below. For example, we may have values such as

$g (- 1.01) = - 4.01$
$g (- 1.001) = - 4.001$

These values are all slightly less than $- 4$ , meaning the outer $g$ is evaluated at inputs to the left of $- 4$ . From the graph, the left-hand limit is

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

2. From the right:

As $x \to - 1^{+}$ , $g (x)$ approaches $- 4$ from above. For example, we might have $g (- 0.99) = - 3.99$ .

These values are all slightly greater $- 4$ , meaning the outer $g$ is evaluated at inputs to the right $- 4$ . From the graph, the right-hand limit is

$x \to - 4^{+} lim g (x) = - 2$

Because the one-sided limits of $g$ as $x$ approaches $- 4$ differ, the one-sided limits of $g (g (x))$ as $x$ approaches $- 1$ also differ, so

$x \to - 1 lim g (g (x)) does not exist.$

Achievable AP Calculus AB

1. Limits

1.2. Analytical limits

Our AP Calculus AB course is now in "early access" - get 50% off for a limited time.

Composite functions

8 min read

Font

Discuss

Feedback

What you’ll learn

Composite functions: Find limits of composite functions algebraically.
Graphical composition: Determine composite limits using given graphs.

A composite function is a function inside another function, usually written as $f (g (x))$ or occasionally as $(f \circ g) (x)$ .

Limit of a composite function:

$x \to a lim g (x) = L$

and $f (x)$ is continuous at $x = L$ , then

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

If $f$ is not continuous at $L$ , one-sided limits must be used to analyze the behavior.

So to find the limit of a composite function,

Find the limit $L$ of the inner function $g (x)$ .
Plug that result into the outer function $f$ , provided $f$ is continuous at $L$ .

Example 1: Continuous functions

If $f (x) = 3 x + 1$ and $g (x) = 2 x^{2}$ , evaluate

a) $x \to 0 lim f (g (x))$

b) $x \to 0 lim g (f (x))$

Solutions

a) $x \to 0 lim f (g (x))$

(spoiler)

First, evaluate the limit of the inner function $g (x)$ to find $L$ :

$x \to 0 lim g (x) = x \to 0 lim 2 x^{2} = 0$

Now evaluate the outer function $f$ at $L = 0$ :

$f (0) = 3 (0) + 1 = 1$

b) $x \to 0 lim g (f (x))$

(spoiler)

First, evaluate the limit of the inner function $f (x)$ to find $L$ :

$x \to 0 lim f (x) = x \to 0 lim (3 x + 1) = 1$

Now evaluate the outer function $g$ at $L = 1$ :

$g (1) = 2 (1)^{2} = 2$

Example 2: Discontinuities

The composite-function limit law works when the outer function $f (x)$ is continuous at the value approached by the inner function $g (x)$ , allowing direct substitution into the outer function.

But if $x \to a lim g (x)$ doesn’t exist or if $f (x)$ is not continuous at $x = L$ , use one-sided limits to analyze the behaviors of each function further.

If $f (x) = \frac{1}{x}$ and $g (x) = x^{2} - 9$ , evaluate:

$x \to 3 lim f (g (x))$

Solution

(spoiler)

First, evaluate the limit of the inner function to find $L$ :

$x \to 3 lim g (x) = x \to 3 lim (x^{2} - 9) = 0$

You can follow along with the explanation by graphing the two functions in Desmos. Enter the following lines:

$g (x) f (x) = x^{2} - 9 = \frac{1}{x}$

Now, observe how $g (x)$ behaves as $x \to 3$ from either side.

1. From the left:

As $x \to 3^{-}$ , the graph of $g$ approaches $0$ from below. For example,
- $g (2.9) = - 0.59$
- $g (2.99) = - 0.0599$
Inputting these negative values into $f (x) = \frac{1}{x}$ gives values that decrease without bound toward $- \infty$ . So:

$x \to 3^{-} lim f (g (x)) = - \infty$

2. From the right:

As $x \to 3^{+}$ , the graph of $g (x)$ approaches $0$ from above (e.g. $g (3.1) = 0.61$ ).
Inputting these positive values into $f (x)$ gives values that increase without bound toward $\infty$ . So:

$x \to 3^{+} lim f (g (x)) = \infty$

Because the two one-sided limits are not equal,

$x \to 3 lim f (g (x)) does not exist.$

You can visualize this by typing $f (g (x))$ directly into Desmos and observing the behavior of the composite function to verify the limits found algebraically.

Evaluating composite limits graphically

Problems similar to the multi-part question below have appeared on past AP exams, including some of the trickier but less common cases.

In the graphs shown below, $g (x)$ has a vertical asymptote at $x = 4$ . The function $f (x)$ is defined only for select values within the interval $(- 5, 8]$ .

Composite function limit

Evaluate:

a) $x \to 4 lim f (g (x))$

b) $x \to 1 lim g (f (x))$

c) $x \to - 2 lim f (f (x))$

d) $x \to - 1 lim g (g (x))$

Answers

(spoiler)

a) Does not exist
b) $\infty$
c) $2$
d) Does not exist

Solutions

a) $x \to 4 lim f (g (x))$

(spoiler)

As $x$ approaches $4$ , the inner function $g (x)$ approaches the vertical asymptote, increasing without bound to $\infty$ on both sides.

However, the outer function $f (x)$ is defined only for inputs in the interval $(- 5, 8]$ . For values of $x$ close to $4$ , $g (x)$ lies outside of the domain of $f$ .

For example, if $g (3.99) = 1000$ then $f (g (3.99)) = f (1000)$ , which is not defined.

Since the composite function $f (g (x))$ is not defined for $x$ -values close to $4$ , the limit does not exist.

b) $x \to 1 lim g (f (x))$

(spoiler)

First, evaluate the inner limit to find $L$ :

$x \to 1 lim f (x) = 4$

1. From the left:

As $x \to 1^{-}$ , the graph shows $f (x)$ approaching $4$ from below. For example, we might have values such as

$f (0.99) = 3.99$
$f (0.999) = 3.999$ , etc.

When these values are inputted into $g$ , such as $g (3.99)$ , $g (3.999)$ , etc. the output grows without bound, and

$x \to 4^{-} lim g (x) = \infty$

2. From the right:

Similarly, as $x \to 1^{+}$ , $f$ also approaches $4$ from below, and $x \to 4^{+} lim g (x) = \infty$ .

Therefore, the composite limit is

$x \to 1 lim g (f (x)) = \infty$

c) $x \to - 2 lim f (f (x))$

(spoiler)

First, evaluate the inner limit:

$x \to - 2 lim f (x) = 3$

Since $f$ is not continuous at $x = 3$ , simply substituting $f (3) = - 3$ does not determine the limit. Instead, we need to analyze how $f (x)$ approaches $3$ .

From the graph, as $x \to - 2$ from either side, $f (x)$ approaches $3$ from above. For example, we might have values such as

$f (- 1.99) = 3.01$
$f (- 2.01) = 3.001$

These values are all slightly greater than $3$ , meaning the outer function is evaluated at inputs to the right of $3$ , and the composite function approaches the right-hand limit of $f$ :

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

Therefore, the composite limit is

$x \to - 2 lim f (f (x)) = 2$

d) $x \to - 1 lim g (g (x))$

(spoiler)

First, evaluate the inner limit:

$x \to - 1 lim g (x) = - 4$

Since $g$ is not continuous at $x = - 4$ , we must analyze how $g$ approaches $- 4$ as $x$ approaches $- 1$ , using one-sided limits.

1. From the left:

As $x \to - 1^{-}$ , the inner function $g (x)$ approaches $- 4$ from below. For example, we may have values such as

$g (- 1.01) = - 4.01$
$g (- 1.001) = - 4.001$

These values are all slightly less than $- 4$ , meaning the outer $g$ is evaluated at inputs to the left of $- 4$ . From the graph, the left-hand limit is

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

2. From the right:

As $x \to - 1^{+}$ , $g (x)$ approaches $- 4$ from above. For example, we might have $g (- 0.99) = - 3.99$ .

These values are all slightly greater $- 4$ , meaning the outer $g$ is evaluated at inputs to the right $- 4$ . From the graph, the right-hand limit is

$x \to - 4^{+} lim g (x) = - 2$

Because the one-sided limits of $g$ as $x$ approaches $- 4$ differ, the one-sided limits of $g (g (x))$ as $x$ approaches $- 1$ also differ, so

$x \to - 1 lim g (g (x)) does not exist.$

Composite functions

Form: $f (g (x))$ or $(f \circ g) (x)$ — a function inside another function
Limit rule: $lim_{x \to a} f (g (x)) = f (lim_{x \to a} g (x)) = f (L)$
- Requires $f (x)$ to be continuous at $x = L$

Steps to evaluate composite limits

Step 1: Find the limit of the inner function
Step 2: Plug that result into the outer function

When the standard rule fails

If $lim_{x \to a} g (x)$ puts values outside the domain of $f$ → limit DNE
If $f$ is discontinuous at $L$ , substituting $f (L)$ is not valid — must analyze one-sided behavior instead

One-sided analysis for tricky cases

Determine whether the inner function approaches $L$ from above or below
Use the corresponding one-sided limit of the outer function
If left- and right-hand composite limits differ → overall limit DNE

Composite functions

What you’ll learn

Composite functions: Find limits of composite functions algebraically.
Graphical composition: Determine composite limits using given graphs.

A composite function is a function inside another function, usually written as $f (g (x))$ or occasionally as $(f \circ g) (x)$ .

Limit of a composite function:

$x \to a lim g (x) = L$

and $f (x)$ is continuous at $x = L$ , then

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

If $f$ is not continuous at $L$ , one-sided limits must be used to analyze the behavior.

So to find the limit of a composite function,

Find the limit $L$ of the inner function $g (x)$ .
Plug that result into the outer function $f$ , provided $f$ is continuous at $L$ .

Example 1: Continuous functions

If $f (x) = 3 x + 1$ and $g (x) = 2 x^{2}$ , evaluate

a) $x \to 0 lim f (g (x))$

b) $x \to 0 lim g (f (x))$

Solutions

a) $x \to 0 lim f (g (x))$

(spoiler)

First, evaluate the limit of the inner function $g (x)$ to find $L$ :

$x \to 0 lim g (x) = x \to 0 lim 2 x^{2} = 0$

Now evaluate the outer function $f$ at $L = 0$ :

$f (0) = 3 (0) + 1 = 1$

b) $x \to 0 lim g (f (x))$

(spoiler)

First, evaluate the limit of the inner function $f (x)$ to find $L$ :

$x \to 0 lim f (x) = x \to 0 lim (3 x + 1) = 1$

Now evaluate the outer function $g$ at $L = 1$ :

$g (1) = 2 (1)^{2} = 2$

Example 2: Discontinuities

The composite-function limit law works when the outer function $f (x)$ is continuous at the value approached by the inner function $g (x)$ , allowing direct substitution into the outer function.

But if $x \to a lim g (x)$ doesn’t exist or if $f (x)$ is not continuous at $x = L$ , use one-sided limits to analyze the behaviors of each function further.

If $f (x) = \frac{1}{x}$ and $g (x) = x^{2} - 9$ , evaluate:

$x \to 3 lim f (g (x))$

Solution

(spoiler)

First, evaluate the limit of the inner function to find $L$ :

$x \to 3 lim g (x) = x \to 3 lim (x^{2} - 9) = 0$

You can follow along with the explanation by graphing the two functions in Desmos. Enter the following lines:

$g (x) f (x) = x^{2} - 9 = \frac{1}{x}$

Now, observe how $g (x)$ behaves as $x \to 3$ from either side.

1. From the left:

As $x \to 3^{-}$ , the graph of $g$ approaches $0$ from below. For example,
- $g (2.9) = - 0.59$
- $g (2.99) = - 0.0599$
Inputting these negative values into $f (x) = \frac{1}{x}$ gives values that decrease without bound toward $- \infty$ . So:

$x \to 3^{-} lim f (g (x)) = - \infty$

2. From the right:

As $x \to 3^{+}$ , the graph of $g (x)$ approaches $0$ from above (e.g. $g (3.1) = 0.61$ ).
Inputting these positive values into $f (x)$ gives values that increase without bound toward $\infty$ . So:

$x \to 3^{+} lim f (g (x)) = \infty$

Because the two one-sided limits are not equal,

$x \to 3 lim f (g (x)) does not exist.$

You can visualize this by typing $f (g (x))$ directly into Desmos and observing the behavior of the composite function to verify the limits found algebraically.

Evaluating composite limits graphically

Problems similar to the multi-part question below have appeared on past AP exams, including some of the trickier but less common cases.

In the graphs shown below, $g (x)$ has a vertical asymptote at $x = 4$ . The function $f (x)$ is defined only for select values within the interval $(- 5, 8]$ .

Evaluate:

a) $x \to 4 lim f (g (x))$

b) $x \to 1 lim g (f (x))$

c) $x \to - 2 lim f (f (x))$

d) $x \to - 1 lim g (g (x))$

Answers

(spoiler)

a) Does not exist
b) $\infty$
c) $2$
d) Does not exist

Solutions

a) $x \to 4 lim f (g (x))$

(spoiler)

As $x$ approaches $4$ , the inner function $g (x)$ approaches the vertical asymptote, increasing without bound to $\infty$ on both sides.

However, the outer function $f (x)$ is defined only for inputs in the interval $(- 5, 8]$ . For values of $x$ close to $4$ , $g (x)$ lies outside of the domain of $f$ .

For example, if $g (3.99) = 1000$ then $f (g (3.99)) = f (1000)$ , which is not defined.

Since the composite function $f (g (x))$ is not defined for $x$ -values close to $4$ , the limit does not exist.

b) $x \to 1 lim g (f (x))$

(spoiler)

First, evaluate the inner limit to find $L$ :

$x \to 1 lim f (x) = 4$

1. From the left:

As $x \to 1^{-}$ , the graph shows $f (x)$ approaching $4$ from below. For example, we might have values such as

$f (0.99) = 3.99$
$f (0.999) = 3.999$ , etc.

When these values are inputted into $g$ , such as $g (3.99)$ , $g (3.999)$ , etc. the output grows without bound, and

$x \to 4^{-} lim g (x) = \infty$

2. From the right:

Similarly, as $x \to 1^{+}$ , $f$ also approaches $4$ from below, and $x \to 4^{+} lim g (x) = \infty$ .

Therefore, the composite limit is

$x \to 1 lim g (f (x)) = \infty$

c) $x \to - 2 lim f (f (x))$

(spoiler)

First, evaluate the inner limit:

$x \to - 2 lim f (x) = 3$

Since $f$ is not continuous at $x = 3$ , simply substituting $f (3) = - 3$ does not determine the limit. Instead, we need to analyze how $f (x)$ approaches $3$ .

From the graph, as $x \to - 2$ from either side, $f (x)$ approaches $3$ from above. For example, we might have values such as

$f (- 1.99) = 3.01$
$f (- 2.01) = 3.001$

These values are all slightly greater than $3$ , meaning the outer function is evaluated at inputs to the right of $3$ , and the composite function approaches the right-hand limit of $f$ :

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

Therefore, the composite limit is

$x \to - 2 lim f (f (x)) = 2$

d) $x \to - 1 lim g (g (x))$

(spoiler)

First, evaluate the inner limit:

$x \to - 1 lim g (x) = - 4$

Since $g$ is not continuous at $x = - 4$ , we must analyze how $g$ approaches $- 4$ as $x$ approaches $- 1$ , using one-sided limits.

1. From the left:

As $x \to - 1^{-}$ , the inner function $g (x)$ approaches $- 4$ from below. For example, we may have values such as

$g (- 1.01) = - 4.01$
$g (- 1.001) = - 4.001$

These values are all slightly less than $- 4$ , meaning the outer $g$ is evaluated at inputs to the left of $- 4$ . From the graph, the left-hand limit is

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

2. From the right:

As $x \to - 1^{+}$ , $g (x)$ approaches $- 4$ from above. For example, we might have $g (- 0.99) = - 3.99$ .

These values are all slightly greater $- 4$ , meaning the outer $g$ is evaluated at inputs to the right $- 4$ . From the graph, the right-hand limit is

$x \to - 4^{+} lim g (x) = - 2$

Because the one-sided limits of $g$ as $x$ approaches $- 4$ differ, the one-sided limits of $g (g (x))$ as $x$ approaches $- 1$ also differ, so

$x \to - 1 lim g (g (x)) does not exist.$