What you’ll learn:

Using direct substitution to find limits
Applying limit laws to simplify expressions
Evaluating limits of piecewise-defined functions with one-sided limits
Handling composite functions

Evaluating limits analytically

Generally, graphs and tables are used only when algebraic methods are difficult to apply or to double-check the solutions. Many limits can be found analytically (using algebra or calculus) with straightforward techniques such as direct substitution and knowledge of limit laws. On this page, we’ll find the limits of various functions, including piecewise-defined and composite functions, using basic analytical methods.

Direct substitution

The simplest method for finding a limit is direct substitution. If plugging in $x = a$ into $f (x)$ results in a finite number, that’s the limit. Always do this first for any limit problem.

Examples

1. Evaluate $x \to 3 lim x^{2}$

Directly substituting $x = 3$ into $x^{2}$ ,

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

Give it a try:

2. Evaluate

$x \to \frac{π}{3} lim \frac{tan ( x ) - 3}{x}$

Solution

(spoiler)

Directly substituting $\frac{π}{3}$ everywhere that $x$ appears,

$= \frac{tan ( π /3 ) - 3}{π /3} = \frac{3 - 3}{π /3} = 0$

Limit properties

Direct substitution works because of a few limit laws that use basic operations to break complicated functions into simpler parts.

Let $x \to a lim f (x) = F$ and $x \to a lim g (x) = G$ . Then,

Constant multiple:

$x \to a lim [b f (x)] = b \cdot F$

Sum or difference:

$x \to a lim [f (x) \pm g (x)] = F \pm G$

Product:

$x \to a lim [f (x) \cdot g (x)] = F \cdot G$

Quotient:

$x \to a lim \frac{f ( x )}{g ( x )} = \frac{F}{G}, G \neq = 0$

Power:

$x \to a lim [f (x)]^{n} = F^{n}$

Root:

$x \to a lim n f (x) = n F, n > 0$

Some AP problems expect you to apply these laws without any explicitly-defined functions. For example,

3. Given:

$x \to 5 lim f (x) = 2$

$x \to 5 lim g (x) = 4$

Evaluate:

$x \to 5 lim [f (x) - 3 g (x)]^{2}$

Using the constant multiple, difference rule, and power rules, the limit becomes:

$(x \to 5 lim f (x) - 3 \cdot x \to 5 lim g (x))^{2}$

$= (2 - 3 \cdot 4)^{2} = (2 - 12)^{2} = (- 10)^{2} = 100$

Piecewise-defined functions

The previous section touched on one-sided limits and how the overall limit exists only if both one-sided limits match. Since piecewise functions are defined by different expressions (pieces, or branches) for different intervals of $x$ , one-sided limits must be used to evaluate any limits involving the breakpoint, or the $x$ -value at which the function is broken into separate branches.

Example

Let $f (x)$ be the piecewise function defined by

$f (x) = ⎩ ⎨ ⎧ x^{2} 2 x 4 if x \leq - 1 if - 1 < x \leq 2 if x > 2$

Evaluate:

a) $x \to - 1 lim f (x)$ b) $x \to 2 lim f (x)$ c) $x \to 3 lim f (x)$

Answers

(spoiler)

a) Does not exist (DNE)
b) $4$
c) $4$

Solutions

a) $x \to - 1 lim f (x)$

(spoiler)

$x$ is approaching $- 1$ , which is one breakpoint of the function. Calculating the one-sided limits:

Left-hand limit: $(x \to - 1^{-})$

$f (x)$ is defined as $x^{2}$ for all $x \leq - 1$ so

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

Right-hand limit: $(x \to - 1^{+})$

$f (x)$ is defined as $2 x$ for $- 1 < x \leq 2$ so

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

Since

$x \to - 1^{-} lim f (x) \neq = x \to - 1^{+} lim f (x)$

the overall limit $x \to - 1 lim f (x)$ does not exist (as it does not approach just one specific value).

b) $x \to 2 lim f (x)$

(spoiler)

$x$ is approaching $2$ , which is another breakpoint of the function.

Left-hand limit: $(x \to 2^{-})$

$f (x)$ is defined as $2 x$ for $- 1 < x \leq 2$ so

$x \to 2^{-} lim 2 x = 4$

Right-hand limit: $(x \to 2^{+})$

$f (x)$ is defined as $4$ for $x > 2$ so

$x \to 2^{+} lim 4 = 4$

Both one-sided limits match, so

$x \to 2 lim f (x) = 4$

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

(spoiler)

Since $x$ is approaching $3$ which is not a breakpoint, one-sided limits aren’t necessary. We just have to figure out which piece to use.

Since $x = 3$ falls into the region $x > 2$ , then we use the 3rd branch, $f (x) = 4$ , and

$x \to 3 lim 4 = 4$

Sidenote

Piecewise in Desmos

You can define piecewise functions by using curly braces to denote the pieces, followed by the inequality and corresponding function for each piece.

Let’s use the function we worked with as an example for what to type

$f (x) = ⎩ ⎨ ⎧ x^{2} 2 x 4 if x \leq - 1 if - 1 < x \leq 2 if x > 2$

This would be entered into Desmos as:

$f (x) = {x \leq - 1 : x^{2}, - 1 < x \leq 2 : 2 x, x > 2 : 4}$

Composite functions

What happens when we have a function inside a function - a composite function $(f \circ g) (x)$ , or $f (g (x))$ ? In this case, the following property holds:

If $f (x)$ is continuous at $x = G$ and $x \to a lim g (x) = G$ , then

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

This means we first find the limit of the inner function $g (x)$ and then evaluate the outer function $f$ at that value.

Examples

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

$x \to 0 lim 2 x^{2} = 0$

Then, plug into $f$ :

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

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

(spoiler)

First, evaluate $x \to 0 lim f (x)$ .

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

Then, plug into $g$ :

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

Confirm visually in Desmos by typing $f (x) = 3 x + 1$ and $g (x) = 2 x^{2}$ to define the two functions, and then typing $f (g (x))$ and $g (f (x))$ to check what $y$ -value each composite function approaches as $x \to 0.$

Challenge problems

Note that the limit law for composite functions applies when the outer function $f (x)$ is continuous at the limit value of the inner function $g$ (more on continuity will be covered in later sections).

But what happens if $x \to a lim g (x)$ doesn’t exist, or if $f (x)$ has breaks in the graph?

In some of these cases, finding $x \to a lim f (g (x))$ requires $x \to G lim f (x)$ (which may not be the same as $f (G)$ ), if $x \to a lim g (x) = G$ . Below is a multi-part problem that encompasses the challenging cases you might encounter on the AP exam, although rare:

Shown below are the graphs of $g (x)$ and $f (x)$ :

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) DNE
b) $\infty$ (DNE)
c) $2$
d) DNE

Solutions

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

(spoiler)

First, evaluate $x \to 4 lim g (x)$ .

There is a vertical asymptote at $x = 4$ and both one-sided limits approach $+ \infty$ , which means the limit doesn’t exist in a sense (no finite value).

More than that, however, is the fact that the outer function $f (x)$ only exists in the domain $(- 5, 8]$ as seen in the graph, so the entire composite function also exists only in this region, with additional restrictions from the inner function $g (x)$ .

Therefore there is no value for “ $f (x)$ at $\infty$ ” and the overall limit does not exist.

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

(spoiler)

First, $x \to 1 lim f (x) = 4$ .

Next, it appears that $g (4)$ does not exist because of the asymptote there. It would be sufficient to answer “DNE” as the overall limit.

But let’s reason through this a little more:

Looking at $f (x)$ , as $x$ approaches from the left of $1$ , $f (x)$ approaches $4$ from below
i.e. $f (0.99) \approx 3.99, f (0.999) \approx 3.999$ , etc.

With these sample output values of $f$ as the input values of $g (x)$ , we see that $g (3.99)$ and $g (3.999) \approx (increasingly large values)$ such that

$x \to 4 lim g (x) = \infty$

Repeating this as $x \to 1^{+}$ of $f (x)$ will show the same result.

So the overall limit $x \to 1 lim g (f (x)) = \infty$ but this also means the limit “does not exist as a finite value” anyway, so “DNE” suffices.

You can visualize these ideas more clearly in Desmos by plotting 3 sample functions that mimic the behavior of each near the points of interest:

$f (x) = x + 3$ , since $x \to 1 lim f (x) = 4$
$g (x) = \frac{1}{( x - 4 ) ^{2}}$ , since the graph shoots up to $\infty$ on either side of the vertical asymptote $x = 4$
$g (f (x))$ , to see the vertical asymptote at $x = 1$ such that

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

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

(spoiler)

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

Because of the breaks in the graph of $f (x)$ , however, the answer isn’t simply $f (3) = - 3$ . Notice that as $x$ approaches $- 2$ from either side, $f (x)$ approaches $3$ from above
i.e. $f (- 1.99) \approx 3.01, f (- 2.01) \approx 3.02$ , etc.

Then the output values of $\approx 3.01$ or $3.02$ are the input values for the outer function $f$ , and we have to use the piece of $f (x)$ to the right of $x = 3$ , looking at $f (3.02)$ or $f (3.01)$ and so on, and

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

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

(spoiler)

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

But which piece do we use as $x \to - 4$ of $g (x)$ ?

As $x$ initially approached $- 1$ from the left side, $g (x)$ approached $- 4$ from below. So a sample value might be $g (- 1.01) = - 4.01$ . Then $g (- 4.01)$ uses the piece to the left of $x = - 4$ , with the limit approaching $- 1$ .

On the other hand, as $x$ initially approached $- 1$ from the right side, $g (x)$ approached $- 4$ from above. A sample value might be $g (- 0.99) = - 3.99$ , and $g (- 3.99)$ uses the piece on the right of $x = - 4$ , with the limit approaching $- 2$ .

Because the two one-sided limits of $g (x)$ as $x \to - 4$ , do not match, the overall limit

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

Key points

For composite functions: to find $x \to a lim f (g (x))$ , find $x \to a lim g (x) = G$ and evaluate $f (G)$ (if $f$ is continuous).
If a limit involves the breakpoint of a piecewise function, use one-sided limits.

What you’ll learn:

Using direct substitution to find limits
Applying limit laws to simplify expressions
Evaluating limits of piecewise-defined functions with one-sided limits
Handling composite functions

Evaluating limits analytically

Direct substitution

The simplest method for finding a limit is direct substitution. If plugging in $x = a$ into $f (x)$ results in a finite number, that’s the limit. Always do this first for any limit problem.

Examples

1. Evaluate $x \to 3 lim x^{2}$

Directly substituting $x = 3$ into $x^{2}$ ,

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

Give it a try:

2. Evaluate

$x \to \frac{π}{3} lim \frac{tan ( x ) - 3}{x}$

Solution

(spoiler)

Directly substituting $\frac{π}{3}$ everywhere that $x$ appears,

$= \frac{tan ( π /3 ) - 3}{π /3} = \frac{3 - 3}{π /3} = 0$

Limit properties

Direct substitution works because of a few limit laws that use basic operations to break complicated functions into simpler parts.

Let $x \to a lim f (x) = F$ and $x \to a lim g (x) = G$ . Then,

Constant multiple:

$x \to a lim [b f (x)] = b \cdot F$

Sum or difference:

$x \to a lim [f (x) \pm g (x)] = F \pm G$

Product:

$x \to a lim [f (x) \cdot g (x)] = F \cdot G$

Quotient:

$x \to a lim \frac{f ( x )}{g ( x )} = \frac{F}{G}, G \neq = 0$

Power:

$x \to a lim [f (x)]^{n} = F^{n}$

Root:

$x \to a lim n f (x) = n F, n > 0$

Some AP problems expect you to apply these laws without any explicitly-defined functions. For example,

3. Given:

$x \to 5 lim f (x) = 2$

$x \to 5 lim g (x) = 4$

Evaluate:

$x \to 5 lim [f (x) - 3 g (x)]^{2}$

Using the constant multiple, difference rule, and power rules, the limit becomes:

$(x \to 5 lim f (x) - 3 \cdot x \to 5 lim g (x))^{2}$

$= (2 - 3 \cdot 4)^{2} = (2 - 12)^{2} = (- 10)^{2} = 100$

Piecewise-defined functions

Example

Let $f (x)$ be the piecewise function defined by

$f (x) = ⎩ ⎨ ⎧ x^{2} 2 x 4 if x \leq - 1 if - 1 < x \leq 2 if x > 2$

Evaluate:

a) $x \to - 1 lim f (x)$ b) $x \to 2 lim f (x)$ c) $x \to 3 lim f (x)$

Answers

(spoiler)

a) Does not exist (DNE)
b) $4$
c) $4$

Solutions

a) $x \to - 1 lim f (x)$

(spoiler)

$x$ is approaching $- 1$ , which is one breakpoint of the function. Calculating the one-sided limits:

Left-hand limit: $(x \to - 1^{-})$

$f (x)$ is defined as $x^{2}$ for all $x \leq - 1$ so

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

Right-hand limit: $(x \to - 1^{+})$

$f (x)$ is defined as $2 x$ for $- 1 < x \leq 2$ so

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

Since

$x \to - 1^{-} lim f (x) \neq = x \to - 1^{+} lim f (x)$

the overall limit $x \to - 1 lim f (x)$ does not exist (as it does not approach just one specific value).

b) $x \to 2 lim f (x)$

(spoiler)

$x$ is approaching $2$ , which is another breakpoint of the function.

Left-hand limit: $(x \to 2^{-})$

$f (x)$ is defined as $2 x$ for $- 1 < x \leq 2$ so

$x \to 2^{-} lim 2 x = 4$

Right-hand limit: $(x \to 2^{+})$

$f (x)$ is defined as $4$ for $x > 2$ so

$x \to 2^{+} lim 4 = 4$

Both one-sided limits match, so

$x \to 2 lim f (x) = 4$

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

(spoiler)

Since $x$ is approaching $3$ which is not a breakpoint, one-sided limits aren’t necessary. We just have to figure out which piece to use.

Since $x = 3$ falls into the region $x > 2$ , then we use the 3rd branch, $f (x) = 4$ , and

$x \to 3 lim 4 = 4$

Sidenote

Piecewise in Desmos

You can define piecewise functions by using curly braces to denote the pieces, followed by the inequality and corresponding function for each piece.

Let’s use the function we worked with as an example for what to type

$f (x) = ⎩ ⎨ ⎧ x^{2} 2 x 4 if x \leq - 1 if - 1 < x \leq 2 if x > 2$

This would be entered into Desmos as:

$f (x) = {x \leq - 1 : x^{2}, - 1 < x \leq 2 : 2 x, x > 2 : 4}$

Composite functions

What happens when we have a function inside a function - a composite function $(f \circ g) (x)$ , or $f (g (x))$ ? In this case, the following property holds:

If $f (x)$ is continuous at $x = G$ and $x \to a lim g (x) = G$ , then

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

This means we first find the limit of the inner function $g (x)$ and then evaluate the outer function $f$ at that value.

Examples

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

$x \to 0 lim 2 x^{2} = 0$

Then, plug into $f$ :

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

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

(spoiler)

First, evaluate $x \to 0 lim f (x)$ .

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

Then, plug into $g$ :

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

Challenge problems

But what happens if $x \to a lim g (x)$ doesn’t exist, or if $f (x)$ has breaks in the graph?

Shown below are the graphs of $g (x)$ and $f (x)$ :

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) DNE
b) $\infty$ (DNE)
c) $2$
d) DNE

Solutions

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

(spoiler)

First, evaluate $x \to 4 lim g (x)$ .

There is a vertical asymptote at $x = 4$ and both one-sided limits approach $+ \infty$ , which means the limit doesn’t exist in a sense (no finite value).

Therefore there is no value for “ $f (x)$ at $\infty$ ” and the overall limit does not exist.

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

(spoiler)

First, $x \to 1 lim f (x) = 4$ .

Next, it appears that $g (4)$ does not exist because of the asymptote there. It would be sufficient to answer “DNE” as the overall limit.

But let’s reason through this a little more:

Looking at $f (x)$ , as $x$ approaches from the left of $1$ , $f (x)$ approaches $4$ from below
i.e. $f (0.99) \approx 3.99, f (0.999) \approx 3.999$ , etc.

With these sample output values of $f$ as the input values of $g (x)$ , we see that $g (3.99)$ and $g (3.999) \approx (increasingly large values)$ such that

$x \to 4 lim g (x) = \infty$

Repeating this as $x \to 1^{+}$ of $f (x)$ will show the same result.

So the overall limit $x \to 1 lim g (f (x)) = \infty$ but this also means the limit “does not exist as a finite value” anyway, so “DNE” suffices.

You can visualize these ideas more clearly in Desmos by plotting 3 sample functions that mimic the behavior of each near the points of interest:

$f (x) = x + 3$ , since $x \to 1 lim f (x) = 4$
$g (x) = \frac{1}{( x - 4 ) ^{2}}$ , since the graph shoots up to $\infty$ on either side of the vertical asymptote $x = 4$
$g (f (x))$ , to see the vertical asymptote at $x = 1$ such that

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

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

(spoiler)

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

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

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

(spoiler)

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

But which piece do we use as $x \to - 4$ of $g (x)$ ?

Because the two one-sided limits of $g (x)$ as $x \to - 4$ , do not match, the overall limit

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

Key points

For composite functions: to find $x \to a lim f (g (x))$ , find $x \to a lim g (x) = G$ and evaluate $f (G)$ (if $f$ is continuous).
If a limit involves the breakpoint of a piecewise function, use one-sided limits.

Analytical limits

What you’ll learn:

Evaluating limits analytically

Direct substitution

Examples

Solution

Limit properties

Piecewise-defined functions

Example

Answers

Solutions

Composite functions

Examples

Solutions

Challenge problems

Answers

Solutions

Analytical limits

What you’ll learn:

Evaluating limits analytically

Direct substitution

Examples

Solution

Limit properties

Piecewise-defined functions

Example

Answers

Solutions

Composite functions

Examples

Solutions

Challenge problems

Answers

Solutions