Limits and infinity | Limits | Achievable AP Calculus AB

What you’ll learn:

How to evaluate limits approaching vertical asymptotes (infinite limits)
Evaluate limits at infinity (horizontal asymptotes)
Dominant term analysis for rational functions
Handling limits at infinity involving square roots

Infinite limits & vertical asymptotes

An infinite limit happens when a function grows without bound (toward $+ \infty$ or $- \infty$ ) as the input $x$ approaches a finite number.

We write this as

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

$or$

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

When you see an infinite limit, it usually means the graph is approaching a vertical asymptote. Many problems involve rational functions and one-sided limits, although trigonometric functions can also produce infinite limits.

Sidenote

Constant over zero

If direct substitution gives a non-zero constant in the numerator and $0$ in the denominator, the result is not indeterminate. It signals a vertical asymptote at $x = a$ .

Examples

Evaluate

$x \to 2 lim \frac{1}{x - 2}$

Solution

Direct substitution gives $\frac{1}{0}$ . That’s not an indeterminate form, and it tells us the function is undefined at $x = 2$ (a vertical asymptote).

To decide whether the function goes to $+ \infty$ or $- \infty$ , look at the sign of the denominator on each side of $2$ .

Approaching from the left ( $x < 2$ ), the denominator $x - 2$ is a small negative number, so $\frac{1}{x - 2}$ is a large negative number. A table confirms this:

$x$	$f (x)$
$1.9$	$- 10$
$1.99$	$- 100$
$1.999$	$- 1000$
$1.9999$	$- 10000$

From the left:

As $x \to 2^{-}$ , $f (x)$ decreases without bound, so

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

From the right:

Approaching from the right ( $x > 2$ ), the denominator $x - 2$ is a small positive number, so $\frac{1}{x - 2}$ is a large positive number. Therefore,

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

Because the one-sided limits are different, the two-sided limit does not exist.

Find the vertical asymptotes of $y = \frac{x - 1}{x ^{2} - 3 x + 2}$

Solution

(spoiler)

Answer: $x = 2 only$

A rational function is undefined where its denominator is $0$ , so start by solving

$x^{2} - 3 x + 2 = 0 (x - 1) (x - 2) = 0 x = 1, 2$

Now check what happens at each value.

At $x = 1$ , substituting into the original function gives $\frac{0}{0}$ , which is indeterminate. That usually means a factor cancels, creating a hole rather than a vertical asymptote.

Compute the limit by factoring and canceling:

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

$= x \to 1 lim \frac{( x - 1 )}{Factor & cancel ( x - 1 ) ( x - 2 )}$

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

$= - 1$

So the function approaches $- 1$ as $x \to 1$ . The value $f (1)$ is still undefined, but the graph has a hole at $x = 1$ , not an asymptote.

At $x = 2$ , substituting gives $\frac{1}{0}$ , which indicates a vertical asymptote.

So the only vertical asymptote is $x = 2$ . The graph behaves like $\frac{1}{x - 2}$ , except it also has a hole at $x = 1$ .

Limits at infinity & horizontal asymptotes

Limits at infinity describe what happens to $f (x)$ as $x$ becomes very large in the positive or negative direction (as $x \to \pm \infty$ ). These limits often reveal horizontal asymptotes, which describe the long-range behavior of the function.

If the limit approaches a finite number $L$ , then $y = L$ is a horizontal asymptote.

A horizontal asymptote is described by

$x \to + \infty lim f (x) = L$

$or$

$x \to - \infty lim f (x) = L$

For limits of rational functions as $x \to \pm \infty$ , a quick method is to compare the highest powers (degrees) in the numerator and denominator. There are three cases:

Case 1: Degree of top $<$ bottom: $H . A . = 0$

e.g. $x \to \infty lim \frac{x ^{2} - 1}{x ^{4} + 1} = 0$

Case 2: Degree of top $=$ bottom: $H . A . = ratio of leading coefficients$

e.g. $x \to \infty lim \frac{2 x ^{2} + 1}{3 x ^{2} - 2} = \frac{2}{3}$

Case 3: Degree of top $>$ bottom: $H . A . = + \infty$ or $- \infty$

e.g. $x \to \infty lim \frac{x ^{3} - 1}{2 x + 1} = \infty$

This “compare the degrees” approach is called dominant term analysis. The idea is that when $∣ x ∣$ is very large, the highest-power terms dominate the behavior, and lower-power terms become negligible.

For example,

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

$\approx x \to \infty lim \frac{x ^{3}}{2 x} = x \to \infty lim \frac{x ^{2}}{2} = \infty (DNE)$

Sidenote

About case 3:

Dominant term analysis is most relevant for case 3. The limit can be $+ \infty$ or $- \infty$ depending on the signs of the leading terms.

Examples

Evaluate

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

Solution

Rewrite the denominator mentally in standard form: the highest power present is $x^{4}$ , so the denominator has degree $4$ . The numerator has degree $3$ .

Since top degree $<$ bottom degree, this is case 1, so the limit should be $0$ . Using dominant terms:

$\approx x \to - \infty lim \frac{- x ^{3}}{x ^{4}}$

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

$= 0$

This matches the horizontal asymptote $y = 0$ .

Evaluate

$x \to \infty lim (- 5 + \frac{1}{x ^{2}})$

Solution

As $x \to \infty$ , the fraction $\frac{1}{x ^{2}} \to 0$ , so the whole expression approaches $- 5$ .

If you want to see it using a single rational expression:

$x \to \infty lim (\frac{- 5 x ^{2}}{x ^{2}} + \frac{1}{x ^{2}})$

$= x \to \infty lim \frac{- 5 x ^{2} + 1}{x ^{2}}$

$\approx x \to \infty lim \frac{- 5 x ^{2}}{x ^{2}}$

$= - 5$

This fits case 2: equal degrees, so the limit is the ratio of leading coefficients $\frac{- 5}{1}$ .

Evaluate

$x \to - \infty lim \frac{- 2 x ^{4} + x - 1}{- 5 x ^{3} - x}$

Solution

The numerator has degree $4$ and the denominator has degree $3$ , so this is case 3. Use dominant terms:

$\approx x \to - \infty lim \frac{- 2 x ^{4}}{- 5 x ^{3}}$

$= x \to - \infty lim \frac{2 x}{5}$

$= - \infty (DNE)$

Square roots

Another common situation is a limit at infinity that includes square roots. A reliable strategy is:

Rewrite the expression so you can factor out the highest power of $x$ inside the radical.
Be careful with $x^{2}$ : it equals $∣ x ∣$ , not $x$ .

Let

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

Evaluate:

a) $x \to \infty lim f (x)$
b) $x \to - \infty lim f (x)$

Solutions

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

(spoiler)

Answer: $1$

Direct substitution gives the indeterminate form $\frac{\infty}{\infty}$ .

Inside the square root, the highest power is $x^{2}$ . Factor it out:

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

Separate the radical:

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

The next step is important:

$x^{2} = ∣ x ∣$ (not just $x$ ). So

$x \to \infty lim \frac{x}{( ∣ x ∣ \cdot \frac{3}{x ^{2}} + 1 ) - 3}$

To handle $∣ x ∣$ , use the fact that for $x \to \infty$ , we have $∣ x ∣ = x$ . Then

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

Factor $x$ from the denominator before canceling:

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

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

As $x \to \infty$ , both $\frac{3}{x ^{2}} \to 0$ and $\frac{3}{x} \to 0$ , so

$\approx x \to \infty lim \frac{1}{0 + 1 - 0} = 1$

b) $x \to - \infty lim f (x)$

(spoiler)

Answer: $- 1$

The algebra is the same as in part (a) up to the point where $x^{2}$ becomes $∣ x ∣$ :

$x \to \infty lim \frac{x}{( ∣ x ∣ \cdot \frac{3}{x ^{2}} + 1 ) - 3}$

Now $x \to - \infty$ , so $∣ x ∣ = - x$ . Substitute that piece:

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

Continue by factoring out $x$ in the denominator and simplifying as before. The limit evaluates to $- 1$ .

Graph $f (x)$ in Desmos to confirm visually!

Not all limits at infinity are rational functions. For example, to evaluate

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

start by using the conjugate to rewrite the expression as a rational expression.

Solution

(spoiler)

Answer: 2

Direct substitution gives the indeterminate form $\infty - \infty$ (which is not $0$ ). Multiply by the conjugate:

$x \to \infty lim (x^{2} + 4 x - x) \cdot Multiplied by conjugate \frac{x ^{2} + 4 x + x}{x ^{2} + 4 x + x}$

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

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

Now factor out the highest power of $x$ inside the square root:

$x \to \infty lim \frac{4 x}{x ^{2} ( 1 + \frac{4}{x} ) + x}$

$= x \to \infty lim \frac{4 x}{( x \cdot 1 + \frac{4}{x} ) + x}$

$= x \to \infty lim \frac{4 x}{x \cdot ( 1 + \frac{4}{x} + 1 )}$

As $x \to \infty$ , $\frac{4}{x} \to 0$ , so

$\approx x \to \infty lim \frac{4}{1 + 0 + 1} = 2$

As a challenge problem, find

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

instead. The answer is, interestingly, $+ \infty$ .

Hint:

(spoiler)

While $\frac{4}{x}$ is close to $0$ for very negative $x$ , it’s still a small negative number. Think about how that affects the expression inside the square root and the overall subtraction.

What you’ll learn:

How to evaluate limits approaching vertical asymptotes (infinite limits)
Evaluate limits at infinity (horizontal asymptotes)
Dominant term analysis for rational functions
Handling limits at infinity involving square roots

Infinite limits & vertical asymptotes

An infinite limit happens when a function grows without bound (toward $+ \infty$ or $- \infty$ ) as the input $x$ approaches a finite number.

We write this as

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

$or$

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

Sidenote

Constant over zero

If direct substitution gives a non-zero constant in the numerator and $0$ in the denominator, the result is not indeterminate. It signals a vertical asymptote at $x = a$ .

Examples

Evaluate

$x \to 2 lim \frac{1}{x - 2}$

Solution

Direct substitution gives $\frac{1}{0}$ . That’s not an indeterminate form, and it tells us the function is undefined at $x = 2$ (a vertical asymptote).

To decide whether the function goes to $+ \infty$ or $- \infty$ , look at the sign of the denominator on each side of $2$ .

Approaching from the left ( $x < 2$ ), the denominator $x - 2$ is a small negative number, so $\frac{1}{x - 2}$ is a large negative number. A table confirms this:

$x$	$f (x)$
$1.9$	$- 10$
$1.99$	$- 100$
$1.999$	$- 1000$
$1.9999$	$- 10000$

From the left:

As $x \to 2^{-}$ , $f (x)$ decreases without bound, so

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

From the right:

Approaching from the right ( $x > 2$ ), the denominator $x - 2$ is a small positive number, so $\frac{1}{x - 2}$ is a large positive number. Therefore,

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

Because the one-sided limits are different, the two-sided limit does not exist.

Find the vertical asymptotes of $y = \frac{x - 1}{x ^{2} - 3 x + 2}$

Solution

(spoiler)

Answer: $x = 2 only$

A rational function is undefined where its denominator is $0$ , so start by solving

$x^{2} - 3 x + 2 = 0 (x - 1) (x - 2) = 0 x = 1, 2$

Now check what happens at each value.

At $x = 1$ , substituting into the original function gives $\frac{0}{0}$ , which is indeterminate. That usually means a factor cancels, creating a hole rather than a vertical asymptote.

Compute the limit by factoring and canceling:

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

$= x \to 1 lim \frac{( x - 1 )}{Factor & cancel ( x - 1 ) ( x - 2 )}$

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

$= - 1$

So the function approaches $- 1$ as $x \to 1$ . The value $f (1)$ is still undefined, but the graph has a hole at $x = 1$ , not an asymptote.

At $x = 2$ , substituting gives $\frac{1}{0}$ , which indicates a vertical asymptote.

So the only vertical asymptote is $x = 2$ . The graph behaves like $\frac{1}{x - 2}$ , except it also has a hole at $x = 1$ .

Limits at infinity & horizontal asymptotes

If the limit approaches a finite number $L$ , then $y = L$ is a horizontal asymptote.

A horizontal asymptote is described by

$x \to + \infty lim f (x) = L$

$or$

$x \to - \infty lim f (x) = L$

For limits of rational functions as $x \to \pm \infty$ , a quick method is to compare the highest powers (degrees) in the numerator and denominator. There are three cases:

Case 1: Degree of top $<$ bottom: $H . A . = 0$

e.g. $x \to \infty lim \frac{x ^{2} - 1}{x ^{4} + 1} = 0$

Case 2: Degree of top $=$ bottom: $H . A . = ratio of leading coefficients$

e.g. $x \to \infty lim \frac{2 x ^{2} + 1}{3 x ^{2} - 2} = \frac{2}{3}$

Case 3: Degree of top $>$ bottom: $H . A . = + \infty$ or $- \infty$

e.g. $x \to \infty lim \frac{x ^{3} - 1}{2 x + 1} = \infty$

For example,

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

$\approx x \to \infty lim \frac{x ^{3}}{2 x} = x \to \infty lim \frac{x ^{2}}{2} = \infty (DNE)$

Sidenote

About case 3:

Dominant term analysis is most relevant for case 3. The limit can be $+ \infty$ or $- \infty$ depending on the signs of the leading terms.

Examples

Evaluate

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

Solution

Rewrite the denominator mentally in standard form: the highest power present is $x^{4}$ , so the denominator has degree $4$ . The numerator has degree $3$ .

Since top degree $<$ bottom degree, this is case 1, so the limit should be $0$ . Using dominant terms:

$\approx x \to - \infty lim \frac{- x ^{3}}{x ^{4}}$

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

$= 0$

This matches the horizontal asymptote $y = 0$ .

Evaluate

$x \to \infty lim (- 5 + \frac{1}{x ^{2}})$

Solution

As $x \to \infty$ , the fraction $\frac{1}{x ^{2}} \to 0$ , so the whole expression approaches $- 5$ .

If you want to see it using a single rational expression:

$x \to \infty lim (\frac{- 5 x ^{2}}{x ^{2}} + \frac{1}{x ^{2}})$

$= x \to \infty lim \frac{- 5 x ^{2} + 1}{x ^{2}}$

$\approx x \to \infty lim \frac{- 5 x ^{2}}{x ^{2}}$

$= - 5$

This fits case 2: equal degrees, so the limit is the ratio of leading coefficients $\frac{- 5}{1}$ .

Evaluate

$x \to - \infty lim \frac{- 2 x ^{4} + x - 1}{- 5 x ^{3} - x}$

Solution

The numerator has degree $4$ and the denominator has degree $3$ , so this is case 3. Use dominant terms:

$\approx x \to - \infty lim \frac{- 2 x ^{4}}{- 5 x ^{3}}$

$= x \to - \infty lim \frac{2 x}{5}$

$= - \infty (DNE)$

Square roots

Another common situation is a limit at infinity that includes square roots. A reliable strategy is:

Rewrite the expression so you can factor out the highest power of $x$ inside the radical.
Be careful with $x^{2}$ : it equals $∣ x ∣$ , not $x$ .

Let

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

Evaluate:

a) $x \to \infty lim f (x)$
b) $x \to - \infty lim f (x)$

Solutions

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

(spoiler)

Answer: $1$

Direct substitution gives the indeterminate form $\frac{\infty}{\infty}$ .

Inside the square root, the highest power is $x^{2}$ . Factor it out:

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

Separate the radical:

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

The next step is important:

$x^{2} = ∣ x ∣$ (not just $x$ ). So

$x \to \infty lim \frac{x}{( ∣ x ∣ \cdot \frac{3}{x ^{2}} + 1 ) - 3}$

To handle $∣ x ∣$ , use the fact that for $x \to \infty$ , we have $∣ x ∣ = x$ . Then

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

Factor $x$ from the denominator before canceling:

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

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

As $x \to \infty$ , both $\frac{3}{x ^{2}} \to 0$ and $\frac{3}{x} \to 0$ , so

$\approx x \to \infty lim \frac{1}{0 + 1 - 0} = 1$

b) $x \to - \infty lim f (x)$

(spoiler)

Answer: $- 1$

The algebra is the same as in part (a) up to the point where $x^{2}$ becomes $∣ x ∣$ :

$x \to \infty lim \frac{x}{( ∣ x ∣ \cdot \frac{3}{x ^{2}} + 1 ) - 3}$

Now $x \to - \infty$ , so $∣ x ∣ = - x$ . Substitute that piece:

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

Continue by factoring out $x$ in the denominator and simplifying as before. The limit evaluates to $- 1$ .

Graph $f (x)$ in Desmos to confirm visually!

Not all limits at infinity are rational functions. For example, to evaluate

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

start by using the conjugate to rewrite the expression as a rational expression.

Solution

(spoiler)

Answer: 2

Direct substitution gives the indeterminate form $\infty - \infty$ (which is not $0$ ). Multiply by the conjugate:

$x \to \infty lim (x^{2} + 4 x - x) \cdot Multiplied by conjugate \frac{x ^{2} + 4 x + x}{x ^{2} + 4 x + x}$

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

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

Now factor out the highest power of $x$ inside the square root:

$x \to \infty lim \frac{4 x}{x ^{2} ( 1 + \frac{4}{x} ) + x}$

$= x \to \infty lim \frac{4 x}{( x \cdot 1 + \frac{4}{x} ) + x}$

$= x \to \infty lim \frac{4 x}{x \cdot ( 1 + \frac{4}{x} + 1 )}$

As $x \to \infty$ , $\frac{4}{x} \to 0$ , so

$\approx x \to \infty lim \frac{4}{1 + 0 + 1} = 2$

As a challenge problem, find

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

instead. The answer is, interestingly, $+ \infty$ .

Hint:

(spoiler)

While $\frac{4}{x}$ is close to $0$ for very negative $x$ , it’s still a small negative number. Think about how that affects the expression inside the square root and the overall subtraction.