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Day 2

Limits At Infinity

Limits at Infinity and Horizontal AsymptotesThe ∞ − ∞ Trick: Rationalizing at Infinity

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)=N(x)D(x)f(x) = \frac{N(x)}{D(x)}, comparing the degrees of NN and DD:
 If deg⁡(N)<deg⁡(D)\deg(N) < \deg(D), the HA is y=0y=0.
 If deg⁡(N)=deg⁡(D)\deg(N) = \deg(D), the HA is y=leading coef of Nleading coef of Dy = \frac{\text{leading coef of } N}{\text{leading coef of } D}.
 If deg⁡(N)>deg⁡(D)\deg(N) > \deg(D), there is no HA (the limit is ±∞\pm\infty).
 Separately: lim⁡x→af(x)=±∞  ⟹  \lim_{x \to a} f(x) = \pm\infty \implies vertical asymptote at x=ax=a.
Watch the TikTok ExplanationAsymptotes of Graphs→
Step-by-Step SOP
  1. 1

    Degree Check

    Quickly identify the highest power in the numerator and denominator.
  2. 2

    Coefficient Ratio

    If the degrees match, divide the leading coefficients to get the HA.
  3. 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 Exercises

Example 01Easy
Find the horizontal and vertical asymptotes of f(x)=2x+1x−3f(x) = \frac{2x+1}{x-3}.
Watch the TikTok ExplanationAsymptotes of Graphs→
NEED A HINT?
For the HA, divide numerator and denominator by the highest power of xx. For the VA, find where the denominator is zero.
SHOW DETAILED EXPLANATION

Step 1: Find the Horizontal Asymptote

lim⁡x→∞2+1x1−3x=2+01−0=2  ⟹  \lim_{x \to \infty} \frac{2 + \frac{1}{x}}{1 - \frac{3}{x}} = \frac{2+0}{1-0} = 2 \implies HA: y=2y=2.

Step 2: Find Vertical Asymptote Candidates

Set the denominator to zero: x−3=0  ⟹  x=3x - 3 = 0 \implies x = 3.

Step 3: Verify the Vertical Asymptote

lim⁡x→3+2x+1x−3=∞\lim_{x \to 3^+} \frac{2x+1}{x-3} = \infty, confirming a vertical asymptote at x=3x=3.

Step 4: State Both Results

HA: y=2y=2, VA: x=3x=3.
Example 02Easy
Find the horizontal asymptote of f(x)=4x2−5x+12x2+9f(x) = \frac{4x^2 - 5x + 1}{2x^2 + 9}.
NEED A HINT?
Check the highest power of xx in the numerator and denominator.
SHOW DETAILED EXPLANATION

Step 1: Compare Degrees

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

Step 2: Take the Ratio

The leading coefficients are 4 and 2. Ratio =4/2=2= 4/2 = 2.

Step 3: State the HA

The HA is the line y=2y=2.
Example 03Hard
Evaluate lim⁡x→∞9x2+12x−5\lim_{x \to \infty} \frac{\sqrt{9x^2 + 1}}{2x - 5}.
NEED A HINT?
Remember that x2\sqrt{x^2} acts like ∣x∣|x|. Consider the signs for ∞\infty vs −∞-\infty.
SHOW DETAILED EXPLANATION

Step 1: Analyze the Degree

The numerator effectively has degree 1 (since x2=x\sqrt{x^2} = x for x>0x>0). The denominator has degree 1.

Step 2: Leading Coefficients

Numerator coefficient: 9=3\sqrt{9} = 3. Denominator coefficient: 2.

Step 3: Solve

The limit is 3/23/2.
Example 04Medium
Evaluate lim⁡x→−∞x3+2x2−5\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 DETAILED EXPLANATION

Step 1: Compare Degrees

Degree 3 (top) > degree 2 (bottom). The limit will be ∞\infty or −∞-\infty, and there is no HA.

Step 2: Check Sign

As x→−∞x \to -\infty: numerator x3→−∞x^3 \to -\infty (negative), denominator x2→+∞x^2 \to +\infty (positive).

Step 3: Result

Negative / Positive = Negative, so the limit is −∞-\infty.
Common Pitfalls
  • ⚠
    The Square Root TrapWhen x→−∞x \to -\infty, x2=−x\sqrt{x^2} = -x (not xx), which flips signs and can create a different HA than the x→+∞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\deg(N) = \deg(D) + 1, there's no horizontal asymptote — instead the graph approaches a slanted line y=mx+by = mx+b, where m=lim⁡x→±∞f(x)xm = \lim_{x \to \pm\infty} \frac{f(x)}{x} and b=lim⁡x→±∞(f(x)−mx)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→±∞(⋅−⋅)\lim_{x \to \pm\infty} (\sqrt{\cdot} - \sqrt{\cdot}) or (⋅−polynomial)(\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.
Step-by-Step SOP
  1. 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. 2

    Multiply by the Conjugate

    Turn the subtraction into a single fraction, then apply the usual degree-comparison rules to the result.

Practice Exercises

Example 01Hard
Evaluate lim⁡x→∞(x2+1−x)\lim_{x \to \infty} \left(\sqrt{x^2+1} - x\right).
NEED A HINT?
This is an ∞−∞\infty - \infty form. Multiply by the conjugate x2+1+xx2+1+x\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}+x}.
SHOW DETAILED EXPLANATION

Step 1: Multiply by the Conjugate

(x2+1−x)⋅x2+1+xx2+1+x=(x2+1)−x2x2+1+x=1x2+1+x\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}.

Step 2: Compare Degrees of the New Fraction

The numerator is degree 0 (constant); the denominator behaves like x+x=2xx + x = 2x (degree 1) as x→∞x \to \infty. Bottom-heavy.

Step 3: Conclude

lim⁡x→∞1x2+1+x=0\lim_{x \to \infty} \frac{1}{\sqrt{x^2+1}+x} = 0.
Example 02Hard
Evaluate lim⁡x→−∞(x2+x−x2−x)\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→−∞x \to -\infty, x2=−x\sqrt{x^2} = -x, not xx.
SHOW DETAILED EXPLANATION

Step 1: Multiply by the Conjugate

(x2+x)−(x2−x)x2+x+x2−x=2xx2+x+x2−x\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}}.

Step 2: Factor Out |x| Carefully

Since x→−∞x \to -\infty, x<0x < 0, so x2+x=∣x∣1+1x=−x1+1x\sqrt{x^2+x} = |x|\sqrt{1+\frac{1}{x}} = -x\sqrt{1+\frac{1}{x}} (similarly for the other root).

Step 3: Simplify

2x−x(1+1x+1−1x)=2−(1+1x+1−1x)\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)}.

Step 4: Take the Limit

As x→−∞x \to -\infty, the terms 1x→0\frac{1}{x} \to 0, so this approaches 2−(1+1)=−1\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→−∞x \to -\infty, x2=−x\sqrt{x^2} = -x, not xx — forgetting this sign flip will give you the wrong final answer (see Example 2).
Gary Chang

Gary Chang

Calculus Educator

5+ Years of Calculus Teaching Experience | AP Calculus Specialist. Dedicated to helping students master calculus through step-by-step logic and clear visualizations.

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