Day 2: Limits at Infinity & Asymptotes

Limits at Infinity and Horizontal Asymptotes

Using the degrees of rational functions to determine the behavior of a function at the far ends of the x-axis, without ever needing a table of huge numbers.

Core Theorem

For

f(x) = \frac{N(x)}{D(x)}

, comparing the degrees of

N

and

D

:
If

\deg(N) < \deg(D)

, the HA is

y=0

.
If

\deg(N) = \deg(D)

, the HA is

y = \frac{\text{leading coef of } N}{\text{leading coef of } D}

.
If

\deg(N) > \deg(D)

, there is no HA (the limit is

\pm\infty

).
Separately:

\lim_{x \to a} f(x) = \pm\infty \implies

vertical asymptote at

x=a

Watch the TikTok ExplanationAsymptotes of Graphs→

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Easy

Find the horizontal and vertical asymptotes of

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

How to think about it

Horizontal and vertical asymptotes come from opposite behaviors. The HA is about end behavior — compare the degrees of top and bottom (or divide by the highest power of

x

). The VA is about where the function blows up — set the denominator to zero.

Key idea: For the HA, divide numerator and denominator by the highest power of $x$ . For the VA, find where the denominator is zero.

Full solution

1. Step 1: Find the Horizontal Asymptote

\lim_{x \to \infty} \frac{2 + \frac{1}{x}}{1 - \frac{3}{x}} = \frac{2+0}{1-0} = 2 \implies

HA:

y=2

2. Step 2: Find Vertical Asymptote Candidates

Set the denominator to zero:

x - 3 = 0 \implies x = 3

3. Step 3: Verify the Vertical Asymptote

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

, confirming a vertical asymptote at

x=3

4. Step 4: State Both Results

HA:

y=2

, VA:

x=3

Worked Example 02Easy

Find the horizontal asymptote of

f(x) = \frac{4x^2 - 5x + 1}{2x^2 + 9}

How to think about it

Key idea: Check the highest power of $x$ in the numerator and denominator.

Full solution

1. Step 1: Compare Degrees

Numerator degree is 2. Denominator degree is 2. They are equal.

2. Step 2: Take the Ratio

The leading coefficients are 4 and 2. Ratio

= 4/2 = 2

3. Step 3: State the HA

The HA is the line

y=2

Step-by-Step SOP

1
Degree Check
Quickly identify the highest power in the numerator and denominator.
2
Coefficient Ratio
If the degrees match, divide the leading coefficients to get the HA.
3
Find VAs Separately
Set the denominator equal to zero, then confirm the numerator doesn't also vanish at the same point (which would signal a hole instead).

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Hard

Evaluate

\lim_{x \to \infty} \frac{\sqrt{9x^2 + 1}}{2x - 5}

Need a hint?

Remember that

\sqrt{x^2}

acts like

|x|

. Consider the signs for

\infty

-\infty

Show solution

1. Step 1: Analyze the Degree

The numerator effectively has degree 1 (since

\sqrt{x^2} = x

for

x>0

). The denominator has degree 1.

2. Step 2: Leading Coefficients

Numerator coefficient:

\sqrt{9} = 3

. Denominator coefficient: 2.

3. Step 3: Solve

The limit is

3/2

Practice 02Medium

Evaluate

\lim_{x \to -\infty} \frac{x^3 + 2}{x^2 - 5}

Need a hint?

The function is 'top-heavy'. Check the sign of the infinity.

Show solution

1. Step 1: Compare Degrees

Degree 3 (top) > degree 2 (bottom). The limit will be

\infty

-\infty

, and there is no HA.

2. Step 2: Check Sign

x \to -\infty

: numerator

x^3 \to -\infty

(negative), denominator

x^2 \to +\infty

(positive).

3. Step 3: Result

Negative / Positive = Negative, so the limit is

-\infty

Common Pitfalls

⚠
The Square Root TrapWhen $x \to -\infty$ , $\sqrt{x^2} = -x$ (not $x$ ), which flips signs and can create a different HA than the $x \to +\infty$ case.
⚠
Ignoring Small TermsOnly the highest power matters at infinity. Don't waste time on lower-degree terms — they vanish in the limit.
⚠
Bonus: Oblique (Slant) AsymptotesWhen $\deg(N) = \deg(D) + 1$ , there's no horizontal asymptote — instead the graph approaches a slanted line $y = mx+b$ , where $m = \lim_{x \to \pm\infty} \frac{f(x)}{x}$ and $b = \lim_{x \to \pm\infty} (f(x) - mx)$ . This isn't a core AP Calc AB topic, but it's good intuition for why 'no HA' doesn't always mean 'no pattern' (see the short video: oblique-asymptotes).

The ∞ − ∞ Trick: Rationalizing at Infinity

When a limit at infinity looks like ' $\infty - \infty$ ', dividing by the highest power doesn't directly apply — instead, multiply by the conjugate to turn subtraction into a fraction you CAN analyze.

Core Theorem

For expressions like

\lim_{x \to \pm\infty} (\sqrt{\cdot} - \sqrt{\cdot})

(\sqrt{\cdot} - \text{polynomial})

: multiply and divide by the conjugate to convert the difference into a single fraction, then apply the usual degree-comparison rules.

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Hard

Evaluate

\lim_{x \to \infty} \left(\sqrt{x^2+1} - x\right)

How to think about it

\sqrt{\text{big}} - \text{big}

is indeterminate because both terms run to infinity together. Multiplying by the conjugate turns the difference of roots into a difference of squares, which simplifies to something you can actually divide out.

Key idea: This is an $\infty - \infty$ form. Multiply by the conjugate $\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}+x}$ .

Full solution

1. Step 1: Multiply by the Conjugate

\left(\sqrt{x^2+1}-x\right) \cdot \frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}+x} = \frac{(x^2+1) - x^2}{\sqrt{x^2+1}+x} = \frac{1}{\sqrt{x^2+1}+x}

2. Step 2: Compare Degrees of the New Fraction

The numerator is degree 0 (constant); the denominator behaves like

x + x = 2x

(degree 1) as

x \to \infty

. Bottom-heavy.

3. Step 3: Conclude

\lim_{x \to \infty} \frac{1}{\sqrt{x^2+1}+x} = 0

Step-by-Step SOP

1
Spot the ∞ − ∞ Form
If you see a difference of two terms that each individually blow up, don't substitute directly — plan to rationalize.
2
Multiply by the Conjugate
Turn the subtraction into a single fraction, then apply the usual degree-comparison rules to the result.

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Hard

Evaluate

\lim_{x \to -\infty} \left(\sqrt{x^2+x} - \sqrt{x^2-x}\right)

Need a hint?

Multiply by the conjugate as before, but be extra careful: as

x \to -\infty

\sqrt{x^2} = -x

, not

x

Show solution

1. Step 1: Multiply by the Conjugate

\frac{(x^2+x)-(x^2-x)}{\sqrt{x^2+x}+\sqrt{x^2-x}} = \frac{2x}{\sqrt{x^2+x}+\sqrt{x^2-x}}

2. Step 2: Factor Out |x| Carefully

Since

x \to -\infty

x < 0

, so

\sqrt{x^2+x} = |x|\sqrt{1+\frac{1}{x}} = -x\sqrt{1+\frac{1}{x}}

(similarly for the other root).

3. Step 3: Simplify

\frac{2x}{-x\left(\sqrt{1+\frac{1}{x}}+\sqrt{1-\frac{1}{x}}\right)} = \frac{2}{-\left(\sqrt{1+\frac{1}{x}}+\sqrt{1-\frac{1}{x}}\right)}

4. Step 4: Take the Limit

x \to -\infty

, the terms

\frac{1}{x} \to 0

, so this approaches

\frac{2}{-(1+1)} = -1

Common Pitfalls

⚠
∞ − ∞ Is Not Automatically 0Two things individually going to infinity, subtracted, doesn't tell you the answer — you must rationalize first to see what's actually left over.
⚠
Same Square Root Trap as BeforeAs $x \to -\infty$ , $\sqrt{x^2} = -x$ , not $x$ — forgetting this sign flip will give you the wrong final answer (see Example 2).

Limits At Infinity

Limits at Infinity and Horizontal Asymptotes

Worked Examples

1. Step 1: Find the Horizontal Asymptote

2. Step 2: Find Vertical Asymptote Candidates

3. Step 3: Verify the Vertical Asymptote

4. Step 4: State Both Results

1. Step 1: Compare Degrees

2. Step 2: Take the Ratio

3. Step 3: State the HA

Degree Check

Coefficient Ratio

Find VAs Separately

Practice — Your Turn

1. Step 1: Analyze the Degree

2. Step 2: Leading Coefficients

3. Step 3: Solve

1. Step 1: Compare Degrees

2. Step 2: Check Sign

3. Step 3: Result

The ∞ − ∞ Trick: Rationalizing at Infinity

Worked Examples

1. Step 1: Multiply by the Conjugate

2. Step 2: Compare Degrees of the New Fraction

3. Step 3: Conclude

Spot the ∞ − ∞ Form

Multiply by the Conjugate

Practice — Your Turn

1. Step 1: Multiply by the Conjugate

2. Step 2: Factor Out |x| Carefully

3. Step 3: Simplify

4. Step 4: Take the Limit

Gary Chang