Limits at Infinity: Rational Functions
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 4.6: “Limits at Infinity and Asymptotes” |
| Direct link | https://openstax.org/books/calculus-volume-1/pages/4-6-limits-at-infinity-and-asymptotes |
| Textbook used in class | Stewart, Calculus, Section 3.4: “Limits at Infinity; Horizontal Asymptotes” |
Opening Scenario
Consider the fraction $\dfrac{100}{x}$. When $x = 1$, this equals $100$. When $x = 1000$, it equals $0.1$. When $x = 1{,}000{,}000$, it equals $0.0001$. As $x$ grows, the fraction shrinks toward $0$. The denominator overwhelms the numerator.
Now consider $\dfrac{x^3}{x}$. As $x$ grows, this simplifies to $x$, which grows without bound. The numerator overwhelms the denominator.
The limit of a rational function as $x \to \infty$ comes down to which polynomial -- numerator or denominator -- grows faster. The technique of dividing by the highest power of $x$ in the denominator makes this comparison exact.
Quick Reference
For a rational function $\dfrac{a_m x^m + \cdots}{b_n x^n + \cdots}$ with $a_m \neq 0$ and $b_n \neq 0$:
| Degrees | $\lim_{x \to \pm\infty} f(x)$ |
|---|---|
| $m < n$ (numerator degree less) | $0$ |
| $m = n$ (equal degrees) | $a_m / b_n$ (ratio of leading coefficients) |
| $m > n$ (numerator degree greater) | $\pm\infty$ (no finite limit) |
Key Concepts
1. The Core Technique: Divide by the Highest Denominator Power
Direct substitution of $\infty$ gives $\infty/\infty$, an indeterminate form. The way out is algebraic: divide every term in both numerator and denominator by $x^n$, where $n$ is the degree of the denominator.
After dividing, every term with $x$ in the denominator has the form $c / x^k$ with $k \geq 1$. As $x \to \infty$, these all go to $0$. What remains is the ratio of leading coefficients.
Example 1. Evaluate $\displaystyle\lim_{x \to \infty} \frac{4x^2 - x + 3}{2x^2 + 7}$.
Highest power in denominator: $x^2$. Divide every term by $x^2$: $$\frac{4x^2/x^2 - x/x^2 + 3/x^2}{2x^2/x^2 + 7/x^2} = \frac{4 - 1/x + 3/x^2}{2 + 7/x^2}.$$
As $x \to \infty$: $1/x \to 0$, $3/x^2 \to 0$, $7/x^2 \to 0$: $$\frac{4 - 0 + 0}{2 + 0} = 2.$$
Boxed answer: $\displaystyle\lim_{x \to \infty} \frac{4x^2 - x + 3}{2x^2 + 7} = 2$.
Recap. The answer $2$ is the ratio of leading coefficients $4/2 = 2$. The technique turns an indeterminate form into a clean arithmetic computation.
2. Numerator Degree Less: Limit is Zero
When the denominator polynomial has higher degree, it grows faster and overwhelms the numerator.
Example 2. $\displaystyle\lim_{x \to \infty} \frac{x + 1}{x^3 - 2}$.
Divide by $x^3$: $$\frac{1/x^2 + 1/x^3}{1 - 2/x^3} \to \frac{0 + 0}{1 - 0} = 0.$$
Boxed answer: $0$.
Recall. This corresponds to “the denominator wins the race to infinity.” Any polynomial grows faster than a polynomial of lower degree.
3. Numerator Degree Greater: Limit is Infinite
When the numerator has higher degree, it grows faster and the ratio grows without bound.
Example 3. $\displaystyle\lim_{x \to \infty} \frac{x^3 + 2}{x + 1}$.
Divide by $x$ (degree of denominator): $$\frac{x^2 + 2/x}{1 + 1/x} \to \frac{x^2}{1} = x^2 \to \infty.$$
Boxed answer: $+\infty$.
The limit is not a real number, so there is no horizontal asymptote in this case.
4. Handling Negative Inputs
As $x \to -\infty$, the technique is the same, but care is needed with square roots and odd powers.
Example 4. $\displaystyle\lim_{x \to -\infty} \frac{2x^2 + x}{x^3 - 1}$.
Divide by $x^3$: $$\frac{2/x + 1/x^2}{1 - 1/x^3}.$$
As $x \to -\infty$: $2/x \to 0$, $1/x^2 \to 0$, $1/x^3 \to 0$.
$$\frac{0 + 0}{1 - 0} = 0.$$
Boxed answer: $0$.
No surprises here since $x^3$ preserves sign and the division works the same way.
5. A Square Root Requires Sign Attention
Example 5. $\displaystyle\lim_{x \to -\infty} \frac{\sqrt{4x^2 + 1}}{x - 2}$.
Factor $\sqrt{x^2}$ from the square root. For large $|x|$, $\sqrt{4x^2 + 1} \approx \sqrt{4x^2} = 2|x|$.
When $x \to -\infty$: $|x| = -x$ (since $x$ is negative). So $\sqrt{4x^2 + 1} \approx -2x$ for large negative $x$.
Formally, divide numerator and denominator by $x$ (which is negative): $$\frac{\sqrt{4x^2+1}/x}{(x-2)/x} = \frac{-\sqrt{4x^2+1}/|x|}{1 - 2/x} = \frac{-\sqrt{4 + 1/x^2}}{1 - 2/x}.$$
As $x \to -\infty$: $\to \dfrac{-\sqrt{4}}{1} = -2$.
Boxed answer: $-2$.
“$\sqrt{x^2} = x$.” For all real $x$, $\sqrt{x^2} = |x|$. When $x < 0$, $|x| = -x$. Forgetting this sign is the single most common error on rational limits involving square roots.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Canceling terms instead of dividing | $\frac{4x^2 - x}{2x^2 + 7}$: “the $x^2$ cancels to give $4/2$” without showing work | Use the division technique; the shortcut gives the same answer but the technique is needed for non-obvious cases |
| Treating $\sqrt{x^2} = x$ for negative $x$ | Getting $+2$ for Example 5 instead of $-2$ | $\sqrt{x^2} = |x|$; when $x < 0$, $|x| = -x$, which flips the sign |
| Concluding a limit of $\infty$ means “no answer” | Writing “DNE” when numerator degree is larger | The limit equals $+\infty$ or $-\infty$ (determine which); it does not exist as a real number, but the behavior is meaningful |
Leveled Practice
Level 1 -- Direct Application
Problem 1. Use the degree-comparison shortcut to state (without detailed computation) the limit of $\dfrac{7x^3 - 2x}{3x^3 + x^2 + 1}$ as $x \to \infty$.
Show answer
Equal degrees ($m = n = 3$). Ratio of leading coefficients: $7/3$.
Boxed answer: $7/3$.
Problem 2. Evaluate $\displaystyle\lim_{x \to \infty} \frac{5 - x^2}{x^2 + x + 1}$.
Show answer
Divide by $x^2$: $$\frac{5/x^2 - 1}{1 + 1/x + 1/x^2} \to \frac{0 - 1}{1 + 0 + 0} = -1.$$
Boxed answer: $-1$.
Level 2 -- Multiple Steps
Problem 3. Evaluate $\displaystyle\lim_{x \to \infty} \frac{\sqrt{9x^2 + x}}{2x + 1}$.
Show answer
As $x \to +\infty$, $x > 0$, so $\sqrt{9x^2 + x} = x\sqrt{9 + 1/x}$.
$$\frac{x\sqrt{9 + 1/x}}{x(2 + 1/x)} = \frac{\sqrt{9 + 1/x}}{2 + 1/x} \to \frac{\sqrt{9}}{2} = \frac{3}{2}.$$
Boxed answer: $3/2$.
Problem 4. Evaluate $\displaystyle\lim_{x \to -\infty} \frac{\sqrt{9x^2 + x}}{2x + 1}$.
Show answer
For $x \to -\infty$, $x < 0$, so $\sqrt{9x^2 + x} = |x|\sqrt{9 + 1/x} = -x\sqrt{9 + 1/x}$.
$$\frac{-x\sqrt{9 + 1/x}}{x(2 + 1/x)} = \frac{-\sqrt{9 + 1/x}}{2 + 1/x} \to \frac{-3}{2}.$$
Boxed answer: $-3/2$.
Contrast with Problem 3: the limits in the two directions are $+3/2$ and $-3/2$, so this function has two horizontal asymptotes.
Mastery Checklist
Mental Model
Think of the limit as a race between the numerator and denominator to reach infinity. The one with the higher degree wins by growing faster. If the denominator wins, the fraction shrinks to $0$ (no horizontal asymptote above zero). If they tie, the fractions cancel to the ratio of the lead runners’ speeds (the leading coefficients). If the numerator wins, the fraction grows without bound (no horizontal asymptote).
The “divide by highest denominator power” technique is just a way to make the race official: it forces the winning term to stay at $1$ while the losing terms shrink to $0$.
Back to Applications of Differentiation | Next: Horizontal Asymptotes