Limits At Infinity for MATH 140
Exam Relevance for MATH 140
Limits at infinity (horizontal asymptotes) are tested heavily in MATH 140. Compare growth rates of polynomials, exponentials, logs.
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What is a Limit at Infinity?
A limit at infinity describes what happens to a function as $x$ grows arbitrarily large (positive or negative).
Notation:
- $\displaystyle\lim_{x \to \infty} f(x) = L$ means $f(x)$ approaches $L$ as $x$ gets larger and larger
- $\displaystyle\lim_{x \to -\infty} f(x) = L$ means $f(x)$ approaches $L$ as $x$ gets more and more negative
🔑 Don't confuse these two concepts:
- Infinite Limits (Skill 90): $x \to a$ but $f(x) \to \pm\infty$ — the output blows up
- Limits at Infinity (this skill): $x \to \pm\infty$ but $f(x) \to L$ — the input goes to infinity
Horizontal Asymptotes
When $\displaystyle\lim_{x \to \infty} f(x) = L$ or $\displaystyle\lim_{x \to -\infty} f(x) = L$, the line $y = L$ is called a horizontal asymptote.
The graph approaches this horizontal line as $x$ goes far to the right or far to the left.
In this graph, we see $f(x) = \frac{1}{x}$. Notice how the function approaches the horizontal line $y = 0$ (the x-axis) as $x$ goes to $+\infty$ or $-\infty$. The function gets closer and closer to $0$ but never actually reaches it.
Key Principle: Powers of $x$ in the Denominator
The fundamental building block:
$$\lim_{x \to \infty} \frac{1}{x^n} = 0 \quad \text{for any } n > 0$$
As $x$ gets huge, $\frac{1}{x^n}$ gets tiny. This is the key fact for evaluating most limits at infinity.
Rational Functions: The Degree Comparison Method
For $\displaystyle\lim_{x \to \pm\infty} \frac{P(x)}{Q(x)}$ where $P$ and $Q$ are polynomials:
Compare the degrees:
| Condition | Result |
|---|---|
| deg(P) < deg(Q) | Limit = $0$ |
| deg(P) = deg(Q) | Limit = $\dfrac{\text{leading coefficient of } P}{\text{leading coefficient of } Q}$ |
| deg(P) > deg(Q) | Limit = $\pm\infty$ (DNE as a finite number) |
The "Divide by Highest Power" Technique
Strategy: Divide both numerator and denominator by the highest power of $x$ in the denominator.
This converts the problem into something with $\frac{1}{x^n}$ terms that go to $0$.
Problem: Find $\displaystyle\lim_{x \to \infty} \frac{3x + 1}{x^2 + 5}$
Step 1: Identify degrees. Numerator: degree 1. Denominator: degree 2.
Since deg(numerator) < deg(denominator), we expect the limit to be $0$.
Step 2: Divide everything by $x^2$ (highest power in denominator):
$$\lim_{x \to \infty} \frac{3x + 1}{x^2 + 5} = \lim_{x \to \infty} \frac{\frac{3x}{x^2} + \frac{1}{x^2}}{\frac{x^2}{x^2} + \frac{5}{x^2}} = \lim_{x \to \infty} \frac{\frac{3}{x} + \frac{1}{x^2}}{1 + \frac{5}{x^2}}$$
Step 3: As $x \to \infty$: $\frac{3}{x} \to 0$, $\frac{1}{x^2} \to 0$, $\frac{5}{x^2} \to 0$
$$= \frac{0 + 0}{1 + 0} = \boxed{0}$$
Problem: Find $\displaystyle\lim_{x \to \infty} \frac{4x^2 - x + 3}{2x^2 + 7x}$
Step 1: Both numerator and denominator have degree 2.
Step 2: Divide everything by $x^2$:
$$\lim_{x \to \infty} \frac{4x^2 - x + 3}{2x^2 + 7x} = \lim_{x \to \infty} \frac{4 - \frac{1}{x} + \frac{3}{x^2}}{2 + \frac{7}{x}}$$
Step 3: As $x \to \infty$, all the $\frac{1}{x}$ and $\frac{1}{x^2}$ terms vanish:
$$= \frac{4 - 0 + 0}{2 + 0} = \boxed{2}$$
Shortcut: When degrees are equal, just take the ratio of leading coefficients: $\frac{4}{2} = 2$
Problem: Find $\displaystyle\lim_{x \to \infty} \frac{x^3 + 2x}{5x^2 - 1}$
Step 1: Numerator degree 3 > Denominator degree 2.
Step 2: Divide by $x^2$:
$$\lim_{x \to \infty} \frac{x^3 + 2x}{5x^2 - 1} = \lim_{x \to \infty} \frac{x + \frac{2}{x}}{5 - \frac{1}{x^2}}$$
Step 3: As $x \to \infty$:
$$= \frac{\infty + 0}{5 - 0} = \boxed{+\infty}$$
The limit does not exist as a finite number—the function grows without bound.
Problem: Find $\displaystyle\lim_{x \to -\infty} \frac{2x - 5}{x + 3}$
Step 1: Same degree (both degree 1).
Step 2: Divide by $x$:
$$\lim_{x \to -\infty} \frac{2x - 5}{x + 3} = \lim_{x \to -\infty} \frac{2 - \frac{5}{x}}{1 + \frac{3}{x}}$$
Step 3: As $x \to -\infty$: $\frac{5}{x} \to 0$ and $\frac{3}{x} \to 0$
$$= \frac{2 - 0}{1 + 0} = \boxed{2}$$
Note: For rational functions, limits at $+\infty$ and $-\infty$ are usually the same!
Problem: Find $\displaystyle\lim_{x \to \infty} \frac{\sqrt{4x^2 + 1}}{3x - 2}$
Step 1: The square root contains $x^2$, so $\sqrt{4x^2 + 1} \approx \sqrt{4x^2} = 2|x| = 2x$ for large positive $x$.
Step 2: Divide numerator and denominator by $x$:
$$\lim_{x \to \infty} \frac{\sqrt{4x^2 + 1}}{3x - 2} = \lim_{x \to \infty} \frac{\frac{\sqrt{4x^2 + 1}}{x}}{\frac{3x - 2}{x}} = \lim_{x \to \infty} \frac{\sqrt{\frac{4x^2 + 1}{x^2}}}{3 - \frac{2}{x}}$$
Step 3: Simplify inside the square root:
$$= \lim_{x \to \infty} \frac{\sqrt{4 + \frac{1}{x^2}}}{3 - \frac{2}{x}} = \frac{\sqrt{4 + 0}}{3 - 0} = \frac{2}{3}$$
$$\boxed{\frac{2}{3}}$$
⚠️ Warning for $x \to -\infty$: When $x < 0$, we have $\sqrt{x^2} = |x| = -x$. This can change the sign of the answer!
Common Mistakes and Misunderstandings
❌ Mistake: Forgetting that $\frac{1}{x} \to 0$ as $x \to \infty$
Wrong: Treating $\frac{1}{x}$ as if it stays significant when $x$ is huge.
Why it's wrong: Any constant divided by an ever-growing quantity shrinks toward zero.
Correct: $\displaystyle\lim_{x \to \infty} \frac{1}{x} = 0$, and similarly $\frac{5}{x} \to 0$, $\frac{1}{x^2} \to 0$, etc.
❌ Mistake: Treating $\frac{\infty}{\infty}$ as $1$
Wrong: $\displaystyle\lim_{x \to \infty} \frac{x^2}{x^2 + 1} = \frac{\infty}{\infty} = 1$
Why it's wrong: $\frac{\infty}{\infty}$ is an indeterminate form—it tells you nothing! You must simplify first.
Correct: Divide by $x^2$: $\displaystyle\lim_{x \to \infty} \frac{1}{1 + \frac{1}{x^2}} = \frac{1}{1 + 0} = 1$ (This one happens to be $1$, but that's not always true!)
❌ Mistake: Forgetting that $\sqrt{x^2} = |x|$, not just $x$
Wrong: For $\displaystyle\lim_{x \to -\infty} \frac{\sqrt{x^2}}{x}$, writing $\frac{x}{x} = 1$.
Why it's wrong: When $x < 0$, we have $\sqrt{x^2} = |x| = -x$, not $x$.
Correct: $\displaystyle\lim_{x \to -\infty} \frac{\sqrt{x^2}}{x} = \frac{-x}{x} = -1$
Quick Reference Summary
| Situation | Strategy | Result |
|---|---|---|
| $\frac{\text{lower degree}}{\text{higher degree}}$ | Divide by highest power | $0$ |
| $\frac{\text{same degree}}{\text{same degree}}$ | Ratio of leading coefficients | $\frac{a_n}{b_n}$ |
| $\frac{\text{higher degree}}{\text{lower degree}}$ | Divide by highest power | $\pm\infty$ |
| Contains $\sqrt{x^2 + \ldots}$ | Factor out $x^2$ from under radical | Watch sign of $x$! |
Basic Limit at Infinity
For any positive power n, 1 divided by x^n approaches 0 as x grows without bound.
Variables:
- $n$:
- any positive exponent
Degree Comparison: Numerator < Denominator
When the numerator polynomial has lower degree than the denominator, the limit is 0.
Variables:
- $P(x)$:
- numerator polynomial
- $Q(x)$:
- denominator polynomial
Degree Comparison: Equal Degrees
When numerator and denominator have the same degree, the limit equals the ratio of leading coefficients.
Variables:
- $a_n$:
- leading coefficient of numerator
- $b_n$:
- leading coefficient of denominator
Degree Comparison: Numerator > Denominator
When the numerator polynomial has higher degree than the denominator, the limit is infinite (DNE as a finite number).
Variables:
- $P(x)$:
- numerator polynomial
- $Q(x)$:
- denominator polynomial
Square Root at Infinity
When simplifying square roots for limits at infinity, remember that √(x²) = |x|, which equals x for x→+∞ but equals -x for x→-∞.
Variables:
- $x$:
- the variable approaching infinity
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