Achievable AP Calculus AB

1. Limits

1.4. Limits and infinity

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Horizontal asymptotes

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What you’ll learn

How to evaluate limits at infinity (horizontal asymptotes) using dominant term analysis
Handling limits at infinity involving square roots

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 end behavior of the function.

By definition, $y = L$ is a horizontal asymptote of $f (x)$ (where $L$ is finite) if, as $x$ approaches infinity,

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

For limits of rational functions as $x \to \pm \infty$ , direct substitution often results in the indeterminate form $\frac{\infty}{\infty}$ . A quick method to evaluate these limits is dominant term analysis, which compares the highest powers (degrees) in the numerator and denominator. The function behaves according to which term grows the fastest, leading to three cases:

Dominant term analysis:

For a rational function $h (x) = \frac{f ( x )}{g ( x )}$ ,

Case 1

If the degree of $f (x) < g (x)$ , then the horizontal asymptote is $y = 0$ , e.g.

$x \to \infty lim \frac{x ^{3} - 1}{x ^{4} + 1} = 0$

Case 2

If the degree of $f (x) = g (x)$ , then the horizontal asymptote is the ratio of the leading coefficients, e.g.

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

Case 3

If the degree of $f (x) > g (x)$ , then there is no horizontal asymptote since $h (x)$ increases or decreases without bound. The limit is either $\infty$ or $- \infty$ , e.g.

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

The idea is that when $x$ has a large magnitude, the highest-power terms dominate the behavior, and lower-power terms become negligible.

The formal algebraic method behind the shortcut is to divide every term by the highest power of $x$ in the denominator and simplify.

This approach is most useful for limits that fit case $3$ , where the resulting infinity depends on how the leading coefficients and degrees interact. For example,

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

First, divide by the highest power of $x$ in the denominator, which is $x^{2}$ , and simplify:

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

Then as $x$ approaches $- \infty$ , both $\frac{1}{x}$ and $\frac{1}{x ^{2}}$ approach $0$ , so the expression behaves like

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

Since $x^{3}$ approaches $- \infty$ as $x \to - \infty$ , the negative signs cancel and

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

Examples

Case 1. Evaluate

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

(spoiler)

Although not written in standard form, the polynomial in the numerator has a degree of $3$ while the denominator has a degree of $4$ .

Since the top degree < bottom degree, this fits case 1, so the limit should be $0$ . Essentially, for a large magnitude of $x$ , the expression can be reduced to the highest powers in the numerator and denominator:

$x \to - \infty lim \frac{- x ^{3}}{x ^{4}} = x \to - \infty lim - \frac{1}{x} = 0$

Graphing the function confirms the horizontal asymptote to be $y = 0$ .

Case 2. Find the horizontal asymptote(s) of

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

(spoiler)

To find horizontal asymptote(s), evaluate both $x \to \infty lim f (x)$ and $x \to - \infty lim f (x)$ .

Both the numerator and the denominator have a degree of $3$ , so the limit to both infinities is just the ratio of the leading coefficients.

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

Therefore the only horizontal asymptote is $y = \frac{1}{2}$ .

Case 3. Evaluate

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

(spoiler)

The numerator has a degree of $4$ and the denominator has a degree of $3$ (case 3). For a large magnitude of $x$ , the expression can be reduced to

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

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)

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:

Since $x^{2} = ∣ x ∣$ ,

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

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

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

Factor $x$ in 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 $\frac{3}{x ^{2}}$ and $\frac{3}{x}$ approach $0$ , so

$\frac{1}{0 + 1 - 0} = 1$

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

(spoiler)

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

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

Because this problem has $x \to - \infty$ , use $∣ x ∣ = - x$ . Substitute that piece:

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

Factor $- x$ from the denominator:

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

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

$\frac{1}{- ( 0 + 1 + 0 )} = - 1$

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

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

use the conjugate to rewrite the expression as a rational expression.

(spoiler)

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

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

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

$\frac{4}{1 + 0 + 1} = 2$

Limits at infinity & horizontal asymptotes

Describe end behavior as $x \to \pm \infty$
Horizontal asymptote: $lim_{x \to \pm \infty} f (x) = L ⟹ y = L$
Dominant term analysis: compare highest degree terms in numerator & denominator

Dominant term analysis for rational functions

Degree numerator $<$ degree denominator: limit $= 0$ (asymptote $y = 0$ )
Degree numerator $=$ degree denominator: limit $=$ ratio of leading coefficients
Degree numerator $>$ degree denominator: limit $\to \pm \infty$ (no horizontal asymptote)

Algebraic method for limits at infinity

Divide all terms by highest power of $x$ in denominator
Simplify and analyze behavior as $x \to \pm \infty$
Useful for determining sign and value in case 3

Square roots in limits at infinity

Factor out highest power of $x$ inside the radical
$x^{2} = ∣ x ∣$ ; use $∣ x ∣ = x$ for $x \to \infty$ , $∣ x ∣ = - x$ for $x \to - \infty$
Simplify and evaluate after factoring

Special techniques for indeterminate forms

For $\infty - \infty$ with radicals: multiply by conjugate to convert to rational form
Factor and cancel highest powers, then take limit

Key examples

Rational function with degree numerator $<$ denominator: limit $= 0$
Rational function with equal degrees: limit $=$ ratio of leading coefficients
Rational function with degree numerator $>$ denominator: limit $\to \pm \infty$
Square root expressions: factor, use $∣ x ∣$ , and simplify
Use conjugate for $x^{2} + b x - x$ type expressions to resolve $\infty - \infty$

Horizontal asymptotes

What you’ll learn

How to evaluate limits at infinity (horizontal asymptotes) using dominant term analysis
Handling limits at infinity involving square roots

By definition, $y = L$ is a horizontal asymptote of $f (x)$ (where $L$ is finite) if, as $x$ approaches infinity,

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

Dominant term analysis:

For a rational function $h (x) = \frac{f ( x )}{g ( x )}$ ,

Case 1

If the degree of $f (x) < g (x)$ , then the horizontal asymptote is $y = 0$ , e.g.

$x \to \infty lim \frac{x ^{3} - 1}{x ^{4} + 1} = 0$

Case 2

If the degree of $f (x) = g (x)$ , then the horizontal asymptote is the ratio of the leading coefficients, e.g.

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

Case 3

If the degree of $f (x) > g (x)$ , then there is no horizontal asymptote since $h (x)$ increases or decreases without bound. The limit is either $\infty$ or $- \infty$ , e.g.

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

The idea is that when $x$ has a large magnitude, the highest-power terms dominate the behavior, and lower-power terms become negligible.

The formal algebraic method behind the shortcut is to divide every term by the highest power of $x$ in the denominator and simplify.

This approach is most useful for limits that fit case $3$ , where the resulting infinity depends on how the leading coefficients and degrees interact. For example,

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

First, divide by the highest power of $x$ in the denominator, which is $x^{2}$ , and simplify:

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

Then as $x$ approaches $- \infty$ , both $\frac{1}{x}$ and $\frac{1}{x ^{2}}$ approach $0$ , so the expression behaves like

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

Since $x^{3}$ approaches $- \infty$ as $x \to - \infty$ , the negative signs cancel and

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

Examples

Case 1. Evaluate

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

(spoiler)

Although not written in standard form, the polynomial in the numerator has a degree of $3$ while the denominator has a degree of $4$ .

$x \to - \infty lim \frac{- x ^{3}}{x ^{4}} = x \to - \infty lim - \frac{1}{x} = 0$

Graphing the function confirms the horizontal asymptote to be $y = 0$ .

Case 2. Find the horizontal asymptote(s) of

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

(spoiler)

To find horizontal asymptote(s), evaluate both $x \to \infty lim f (x)$ and $x \to - \infty lim f (x)$ .

Both the numerator and the denominator have a degree of $3$ , so the limit to both infinities is just the ratio of the leading coefficients.

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

Therefore the only horizontal asymptote is $y = \frac{1}{2}$ .

Case 3. Evaluate

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

(spoiler)

The numerator has a degree of $4$ and the denominator has a degree of $3$ (case 3). For a large magnitude of $x$ , the expression can be reduced to

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

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)

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:

Since $x^{2} = ∣ x ∣$ ,

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

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

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

Factor $x$ in 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 $\frac{3}{x ^{2}}$ and $\frac{3}{x}$ approach $0$ , so

$\frac{1}{0 + 1 - 0} = 1$

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

(spoiler)

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

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

Because this problem has $x \to - \infty$ , use $∣ x ∣ = - x$ . Substitute that piece:

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

Factor $- x$ from the denominator:

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

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

$\frac{1}{- ( 0 + 1 + 0 )} = - 1$

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

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

use the conjugate to rewrite the expression as a rational expression.

(spoiler)

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

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

$\frac{4}{1 + 0 + 1} = 2$

Key points

Limits at infinity & horizontal asymptotes

Describe end behavior as $x \to \pm \infty$
Horizontal asymptote: $lim_{x \to \pm \infty} f (x) = L ⟹ y = L$
Dominant term analysis: compare highest degree terms in numerator & denominator

Dominant term analysis for rational functions

Degree numerator $<$ degree denominator: limit $= 0$ (asymptote $y = 0$ )
Degree numerator $=$ degree denominator: limit $=$ ratio of leading coefficients
Degree numerator $>$ degree denominator: limit $\to \pm \infty$ (no horizontal asymptote)

Algebraic method for limits at infinity

Divide all terms by highest power of $x$ in denominator
Simplify and analyze behavior as $x \to \pm \infty$
Useful for determining sign and value in case 3

Square roots in limits at infinity

Factor out highest power of $x$ inside the radical
$x^{2} = ∣ x ∣$ ; use $∣ x ∣ = x$ for $x \to \infty$ , $∣ x ∣ = - x$ for $x \to - \infty$
Simplify and evaluate after factoring

Special techniques for indeterminate forms

For $\infty - \infty$ with radicals: multiply by conjugate to convert to rational form
Factor and cancel highest powers, then take limit

Key examples

Rational function with degree numerator $<$ denominator: limit $= 0$
Rational function with equal degrees: limit $=$ ratio of leading coefficients
Rational function with degree numerator $>$ denominator: limit $\to \pm \infty$
Square root expressions: factor, use $∣ x ∣$ , and simplify
Use conjugate for $x^{2} + b x - x$ type expressions to resolve $\infty - \infty$

What you’ll learn

How to evaluate limits at infinity (horizontal asymptotes) using dominant term analysis
Handling limits at infinity involving square roots

By definition, $y = L$ is a horizontal asymptote of $f (x)$ (where $L$ is finite) if, as $x$ approaches infinity,

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

Dominant term analysis:

For a rational function $h (x) = \frac{f ( x )}{g ( x )}$ ,

Case 1

If the degree of $f (x) < g (x)$ , then the horizontal asymptote is $y = 0$ , e.g.

$x \to \infty lim \frac{x ^{3} - 1}{x ^{4} + 1} = 0$

Case 2

If the degree of $f (x) = g (x)$ , then the horizontal asymptote is the ratio of the leading coefficients, e.g.

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

Case 3

If the degree of $f (x) > g (x)$ , then there is no horizontal asymptote since $h (x)$ increases or decreases without bound. The limit is either $\infty$ or $- \infty$ , e.g.

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

The idea is that when $x$ has a large magnitude, the highest-power terms dominate the behavior, and lower-power terms become negligible.

The formal algebraic method behind the shortcut is to divide every term by the highest power of $x$ in the denominator and simplify.

This approach is most useful for limits that fit case $3$ , where the resulting infinity depends on how the leading coefficients and degrees interact. For example,

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

First, divide by the highest power of $x$ in the denominator, which is $x^{2}$ , and simplify:

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

Then as $x$ approaches $- \infty$ , both $\frac{1}{x}$ and $\frac{1}{x ^{2}}$ approach $0$ , so the expression behaves like

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

Since $x^{3}$ approaches $- \infty$ as $x \to - \infty$ , the negative signs cancel and

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

Examples

Case 1. Evaluate

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

(spoiler)

Although not written in standard form, the polynomial in the numerator has a degree of $3$ while the denominator has a degree of $4$ .

$x \to - \infty lim \frac{- x ^{3}}{x ^{4}} = x \to - \infty lim - \frac{1}{x} = 0$

Graphing the function confirms the horizontal asymptote to be $y = 0$ .

Case 2. Find the horizontal asymptote(s) of

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

(spoiler)

To find horizontal asymptote(s), evaluate both $x \to \infty lim f (x)$ and $x \to - \infty lim f (x)$ .

Both the numerator and the denominator have a degree of $3$ , so the limit to both infinities is just the ratio of the leading coefficients.

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

Therefore the only horizontal asymptote is $y = \frac{1}{2}$ .

Case 3. Evaluate

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

(spoiler)

The numerator has a degree of $4$ and the denominator has a degree of $3$ (case 3). For a large magnitude of $x$ , the expression can be reduced to

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

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)

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:

Since $x^{2} = ∣ x ∣$ ,

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

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

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

Factor $x$ in 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 $\frac{3}{x ^{2}}$ and $\frac{3}{x}$ approach $0$ , so

$\frac{1}{0 + 1 - 0} = 1$

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

(spoiler)

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

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

Because this problem has $x \to - \infty$ , use $∣ x ∣ = - x$ . Substitute that piece:

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

Factor $- x$ from the denominator:

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

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

$\frac{1}{- ( 0 + 1 + 0 )} = - 1$

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

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

use the conjugate to rewrite the expression as a rational expression.

(spoiler)

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

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

$\frac{4}{1 + 0 + 1} = 2$