Evaluating Limits of Rational Functions at Infinity

MATH161

Reference: Stewart 3.4 • Chapter: 3 • Section: 4

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Evaluating Limits of Rational Functions at Infinity

The Battle of the Polynomials

When you have a fraction with polynomials in both the numerator and denominator, and you let $x \to \infty$, what happens? Both the top and bottom get large—but who wins?

The answer depends on which polynomial grows faster. This section gives you a systematic technique to evaluate these limits: divide everything by the highest power of $x$ in the denominator. This transforms an indeterminate "$\frac{\infty}{\infty}$" into a clean, computable limit.

This technique is one of the most frequently tested skills in Calculus I, appearing on almost every exam.

Prerequisite Map

Quick Reference

Property	Value
Concept	Limits
Course	MATH161
Section	Stewart 2.6
Difficulty	Intermediate
Time	~25 minutes

Key Concepts

The Divide-by-Highest-Power Technique

The Method: To evaluate $\lim_{x \to \infty} \frac{P(x)}{Q(x)}$ where $P$ and $Q$ are polynomials:

Identify the highest power of $x$ in the denominator
Divide every term (top and bottom) by that power of $x$
Take the limit, using $\lim_{x \to \infty} \frac{1}{x^k} = 0$ for all $k > 0$

Worked Example

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

Step 1: Highest power in denominator is $x^2$

Step 2: Divide every term by $x^2$: $$\frac{3x^2 - x - 2}{5x^2 + 4x + 1} = \frac{\frac{3x^2}{x^2} - \frac{x}{x^2} - \frac{2}{x^2}}{\frac{5x^2}{x^2} + \frac{4x}{x^2} + \frac{1}{x^2}} = \frac{3 - \frac{1}{x} - \frac{2}{x^2}}{5 + \frac{4}{x} + \frac{1}{x^2}}$$

Step 3: Take the limit as $x \to \infty$: $$\lim_{x \to \infty} \frac{3 - \frac{1}{x} - \frac{2}{x^2}}{5 + \frac{4}{x} + \frac{1}{x^2}} = \frac{3 - 0 - 0}{5 + 0 + 0} = \frac{3}{5}$$

The Three Cases: Degree Comparison

Let $\deg(P)$ = degree of numerator, $\deg(Q)$ = degree of denominator.

Case	Condition	Result	Horizontal Asymptote
Denominator wins	$\deg(P) < \deg(Q)$	$\lim = 0$	$y = 0$
Tie	$\deg(P) = \deg(Q)$	$\lim = \frac{a_n}{b_n}$	$y = \frac{\text{leading coeff. of } P}{\text{leading coeff. of } Q}$
Numerator wins	$\deg(P) > \deg(Q)$	$\lim = \pm\infty$	None

Why This Works

When $x$ is very large, only the highest-power terms matter:

$$\frac{3x^2 - x - 2}{5x^2 + 4x + 1} \approx \frac{3x^2}{5x^2} = \frac{3}{5}$$

The lower-power terms become negligible compared to the dominant terms.

Visual Intuition

Degree of numerator < Degree of denominator:
    Denominator grows faster → fraction shrinks → limit is 0

    y     ___________
    |    /           \
    |   /             \_______________  ← approaches 0
    |  /
    +────────────────────────────→ x

Degree of numerator = Degree of denominator:
    Same growth rate → ratio of leading coefficients

    y
    |
  L ├───────────────────────────────  ← horizontal asymptote at L
    |         ~~~~~~~~~~~~~~~~~~~~~~~~
    +────────────────────────────→ x

Degree of numerator > Degree of denominator:
    Numerator grows faster → fraction grows → no horizontal asymptote

    y
    |                          /
    |                       /
    |                    /
    |                 /
    +────────────────────────────→ x

Warning: Infinite Limits at Infinity

When $\deg(P) > \deg(Q)$, the limit is $\pm\infty$. The sign depends on:

The signs of the leading coefficients
Whether $x \to +\infty$ or $x \to -\infty$
Whether the degree difference is odd or even

Example: $\lim_{x \to \infty} \frac{x^3}{x+1}$

The degree of the numerator (3) exceeds the degree of the denominator (1). As $x \to \infty$: $$\frac{x^3}{x+1} \approx \frac{x^3}{x} = x^2 \to \infty$$

So $\lim_{x \to \infty} \frac{x^3}{x+1} = \infty$.

Practice Problems

Level 1 Comparing Degrees

Without computing, determine the limit as $x \to \infty$ for each function based on degree comparison:

$\frac{x^2 + 1}{x^5 - 3x}$
$\frac{7x^4 - 2x}{3x^4 + 5}$
$\frac{x^6}{x^3 + x^2 + x + 1}$

Thought Process

For each fraction, identify the degree of the numerator and denominator. Apply the three-case rule: if numerator degree < denominator degree, limit is 0; if equal, limit is ratio of leading coefficients; if numerator > denominator, limit is $\pm\infty$.

Show Answer

(a) $\frac{x^2 + 1}{x^5 - 3x}$

Numerator degree: 2
Denominator degree: 5
Since $2 < 5$: $\lim_{x \to \infty} = 0$

(b) $\frac{7x^4 - 2x}{3x^4 + 5}$

Numerator degree: 4
Denominator degree: 4
Since degrees are equal: $\lim_{x \to \infty} = \frac{7}{3}$

(c) $\frac{x^6}{x^3 + x^2 + x + 1}$

Numerator degree: 6
Denominator degree: 3
Since $6 > 3$: $\lim_{x \to \infty} = \infty$

Level 2 Applying the Divide-by-Highest-Power Technique

Evaluate $\lim_{x \to \infty} \frac{2x^3 + 5x - 1}{4x^3 - x^2 + 7}$ by dividing by the highest power of $x$ in the denominator.

Thought Process

The highest power in the denominator is $x^3$. Divide every term by $x^3$, then let the $\frac{1}{x^k}$ terms go to zero.

Show Answer

Divide numerator and denominator by $x^3$:

$$\frac{2x^3 + 5x - 1}{4x^3 - x^2 + 7} = \frac{2 + \frac{5}{x^2} - \frac{1}{x^3}}{4 - \frac{1}{x} + \frac{7}{x^3}}$$

As $x \to \infty$:

$\frac{5}{x^2} \to 0$
$\frac{1}{x^3} \to 0$
$\frac{1}{x} \to 0$
$\frac{7}{x^3} \to 0$

Therefore: $$\lim_{x \to \infty} \frac{2 + \frac{5}{x^2} - \frac{1}{x^3}}{4 - \frac{1}{x} + \frac{7}{x^3}} = \frac{2 + 0 - 0}{4 - 0 + 0} = \boxed{\frac{1}{2}}$$

The horizontal asymptote is $y = \frac{1}{2}$.

Level 3 Limit as x → -∞

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

Is this limit the same as the limit as $x \to +\infty$?

Thought Process

The technique works the same for $x \to -\infty$. Divide by the highest power ($x^2$). The key question: do the $\frac{1}{x^k}$ terms still go to zero as $x \to -\infty$? Yes, they do! Then compare with the $x \to +\infty$ case.

Show Answer

Divide by $x^2$: $$\frac{x^2 + 3x}{2x^2 - 5} = \frac{1 + \frac{3}{x}}{2 - \frac{5}{x^2}}$$

As $x \to -\infty$:

$\frac{3}{x} \to 0$ (approaches 0 from negative values)
$\frac{5}{x^2} \to 0$ (still positive, approaching 0)

Therefore: $$\lim_{x \to -\infty} \frac{1 + \frac{3}{x}}{2 - \frac{5}{x^2}} = \frac{1 + 0}{2 - 0} = \boxed{\frac{1}{2}}$$

Comparison with $x \to +\infty$:

$$\lim_{x \to +\infty} \frac{x^2 + 3x}{2x^2 - 5} = \frac{1 + 0}{2 - 0} = \frac{1}{2}$$

Yes, both limits are the same. This happens because the degrees are equal, and even powers dominate. The horizontal asymptote $y = \frac{1}{2}$ applies in both directions.

Note: For rational functions where the degrees are equal, the limits at $+\infty$ and $-\infty$ are always equal (same horizontal asymptote on both sides).

Level 4 Finding All Asymptotes

Find all asymptotes (both vertical and horizontal) of: $$f(x) = \frac{x^2 - 9}{x^2 - 4}$$

Thought Process

For vertical asymptotes: factor and find where the denominator is zero (after canceling common factors). For horizontal asymptotes: use the degree comparison method.

Show Answer

Factoring: $$f(x) = \frac{x^2 - 9}{x^2 - 4} = \frac{(x-3)(x+3)}{(x-2)(x+2)}$$

No common factors, so no holes.

Vertical Asymptotes:

The denominator is zero when $x = 2$ or $x = -2$.

Vertical asymptotes: $x = 2$ and $x = -2$

Horizontal Asymptote:

Degree of numerator = 2, degree of denominator = 2.

Since degrees are equal: $$\lim_{x \to \pm\infty} f(x) = \frac{\text{leading coeff. of numerator}}{\text{leading coeff. of denominator}} = \frac{1}{1} = 1$$

Horizontal asymptote: $y = 1$

Summary:

Vertical asymptotes: $x = 2$, $x = -2$
Horizontal asymptote: $y = 1$
Total: 3 asymptotes

Level 5 General Polynomial Ratio Limit

Let $P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$ with $a_n \neq 0$, and let $Q(x) = b_m x^m + b_{m-1} x^{m-1} + \cdots + b_1 x + b_0$ with $b_m \neq 0$.

Prove that: $$\lim_{x \to \infty} \frac{P(x)}{Q(x)} = \begin{cases} 0 & \text{if } n < m \\ \frac{a_n}{b_m} & \text{if } n = m \\ \pm\infty & \text{if } n > m \end{cases}$$

Thought Process

Divide numerator and denominator by $x^m$ (the highest power in the denominator). Factor out powers of $x$ carefully and analyze what happens to each term.

Show Answer

Proof:

Divide both numerator and denominator by $x^m$:

$$\frac{P(x)}{Q(x)} = \frac{a_n x^{n-m} + a_{n-1} x^{n-1-m} + \cdots + a_0 x^{-m}}{b_m + b_{m-1} x^{-1} + \cdots + b_0 x^{-m}}$$

Case 1: $n < m$

All powers of $x$ in the numerator are negative: $n - m < 0$, $n - 1 - m < 0$, etc.

As $x \to \infty$, every term in the numerator approaches 0 (since $x^{-k} \to 0$ for $k > 0$).

The denominator approaches $b_m \neq 0$.

Therefore: $\lim_{x \to \infty} \frac{P(x)}{Q(x)} = \frac{0}{b_m} = 0$ ✓

Case 2: $n = m$

The leading term in the numerator becomes $a_n x^0 = a_n$.

$$\frac{P(x)}{Q(x)} = \frac{a_n + a_{n-1} x^{-1} + \cdots + a_0 x^{-n}}{b_m + b_{m-1} x^{-1} + \cdots + b_0 x^{-m}}$$

As $x \to \infty$, all $x^{-k}$ terms vanish:

$$\lim_{x \to \infty} \frac{P(x)}{Q(x)} = \frac{a_n + 0 + \cdots + 0}{b_m + 0 + \cdots + 0} = \frac{a_n}{b_m}$$ ✓

Case 3: $n > m$

The leading term has positive power: $a_n x^{n-m}$ where $n - m > 0$.

We can factor: $$\frac{P(x)}{Q(x)} = x^{n-m} \cdot \frac{a_n + a_{n-1} x^{-1} + \cdots}{b_m + b_{m-1} x^{-1} + \cdots}$$

As $x \to \infty$:

$x^{n-m} \to \infty$ (since $n - m > 0$)
The fraction approaches $\frac{a_n}{b_m}$ (a nonzero constant)

The product of something going to infinity and a nonzero constant is $\pm\infty$.

The sign is determined by the sign of $\frac{a_n}{b_m}$ (and whether $x \to +\infty$ or $-\infty$ when $n-m$ is odd).

Therefore: $\lim_{x \to \infty} \frac{P(x)}{Q(x)} = \pm\infty$ ✓

$\blacksquare$

Mastery Checklist

[ ] I can identify the highest power in the denominator of a rational function
[ ] I can divide all terms by that power to simplify the limit
[ ] I know the three cases: degree comparison determines whether limit is 0, finite nonzero, or infinite
[ ] I can quickly read off horizontal asymptotes from the degree/leading coefficient comparison
[ ] I can evaluate limits as $x \to -\infty$ (same technique applies)
[ ] I can find both vertical and horizontal asymptotes of a rational function

Mental Model

The Weight Class Analogy: Think of polynomials like boxers in different weight classes. The degree is the weight class. When two polynomials "fight" (form a ratio):

If the denominator is in a higher weight class (higher degree), it dominates—the ratio goes to 0
If they're in the same weight class, it's a fair fight—the ratio of their leading coefficients decides the outcome
If the numerator is in a higher weight class, it dominates—the ratio goes to infinity

The lower-degree terms are like "reach" or "speed"—they might matter at first, but for large $x$, only the weight class (degree) matters.

Connections

Looking back:

Limits at Infinity Definition explains what these limits mean
Polynomial Functions covers how to identify degree and leading coefficients

Looking ahead:

Special Techniques handles radicals and other non-polynomial cases
Curve Sketching uses these limits to determine end behavior of graphs
L'Hôpital's Rule (Calculus II) provides an alternative method for these limits

Real-world connections:

Chemical reaction rates often have rational function models where the limit represents equilibrium
In networking, packet delivery probability as load increases follows rational models
Population dynamics with limited resources approach carrying capacity ratios

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Limits at Infinity Definition	Skills Index	Special Techniques for Limits at Infinity

Last updated: 2026-01-22