Limits at Infinity and Asymptotes
Explore function behavior as values grow without bound
You have probably noticed that some things level off. A new employee gets faster at their job each week, but eventually they hit a ceiling and cannot really improve much more. A rumor spreads rapidly at first, but once nearly everyone has heard it, the growth slows to a crawl. A hot cup of coffee cools quickly at first, then approaches room temperature more and more slowly, never quite getting there. These situations share something profound: they all involve quantities that approach a limiting value as time (or some other variable) grows without bound.
This is exactly what limits at infinity capture mathematically. When we write $\lim_{x \to \infty} f(x) = L$, we are asking a very natural question: “Where does this function end up heading as we keep going further and further along?” The answer tells us about the long-term behavior of the function, and it turns out this question has enormous practical importance in physics, economics, biology, and engineering.
The good news is that evaluating limits at infinity often boils down to simple comparisons. Which part of the function dominates when numbers get huge? Once you learn to identify the “main character” in a function (read: the term that matters most when values are astronomical), limits at infinity become surprisingly manageable.
Core Concepts
What Does “Infinity” Mean Here?
First, let us be clear: infinity is not a number. You cannot plug infinity into a function and get a result. When we write $x \to \infty$, we are describing a process: $x$ keeps growing, without any upper bound. It goes past 100, past 1,000, past a million, past any number you can name.
So $\lim_{x \to \infty} f(x) = L$ means: “As $x$ gets larger and larger without bound, the values $f(x)$ get closer and closer to $L$.” The function might never actually reach $L$, but it gets arbitrarily close.
Similarly, $x \to -\infty$ means $x$ becomes more and more negative without bound: it goes past $-100$, past $-1000$, and so on, heading down the negative number line forever.
Limits at Infinity: The Basic Idea
Consider the function $f(x) = \frac{1}{x}$. What happens as $x$ gets very large?
| $x$ | $f(x) = \frac{1}{x}$ |
|---|---|
| 10 | 0.1 |
| 100 | 0.01 |
| 1,000 | 0.001 |
| 1,000,000 | 0.000001 |
As $x$ grows, $\frac{1}{x}$ shrinks toward zero. We write:
$$\lim_{x \to \infty} \frac{1}{x} = 0$$
This is one of the most fundamental limits at infinity, and it serves as a building block for many others. The same logic gives us $\lim_{x \to \infty} \frac{1}{x^n} = 0$ for any positive $n$.
Horizontal Asymptotes
When a function approaches a specific value $L$ as $x$ goes to positive or negative infinity, the horizontal line $y = L$ is called a horizontal asymptote.
Think of a horizontal asymptote as a “target line” that the function approaches but may never actually reach. The function might wiggle around it, approach it from above or below, or even cross it multiple times, but as $x$ heads toward infinity, the function values settle toward that line.
A function can have:
- No horizontal asymptote (if the function grows without bound)
- One horizontal asymptote (same behavior as $x \to \infty$ and $x \to -\infty$)
- Two different horizontal asymptotes (different behavior in each direction)
Comparing Growth Rates: The Key Insight
When $x$ is huge, different mathematical expressions grow at vastly different rates. This is the secret to evaluating most limits at infinity: figure out which part of your function “wins” when values get astronomical.
Here is the hierarchy of growth rates, from slowest to fastest:
$$\text{constant} \ll \ln(x) \ll \sqrt{x} \ll x \ll x^2 \ll x^3 \ll e^x \ll x!$$
The symbol $\ll$ means “grows much slower than.” When you have a sum or difference of terms with different growth rates, the fastest-growing term dominates and the others become negligible.
For polynomials specifically, this means: the highest-degree term dominates. In the polynomial $3x^4 + 17x^2 - 500x + 1000$, when $x = 1,000,000$, the $3x^4$ term equals $3 \times 10^{24}$, while the constant term is just $1000$. The highest-degree term completely overwhelms everything else.
Rational Functions: The Degree Comparison Technique
A rational function is a ratio of two polynomials, like $\frac{2x^2 + 3x}{5x^2 - 1}$. For these functions, there is a simple rule based on comparing the degrees of the numerator and denominator:
Case 1: Degree of numerator < Degree of denominator
$$\lim_{x \to \infty} \frac{3x + 1}{x^2 + 5} = 0$$
The denominator grows faster, so the whole fraction shrinks to zero. Horizontal asymptote: $y = 0$.
Case 2: Degree of numerator = Degree of denominator
$$\lim_{x \to \infty} \frac{2x^2 + 3x}{5x^2 - 1} = \frac{2}{5}$$
The limit equals the ratio of the leading coefficients. Horizontal asymptote: $y = \frac{2}{5}$.
Case 3: Degree of numerator > Degree of denominator
$$\lim_{x \to \infty} \frac{x^3 + 2x}{x^2 - 4} = \infty$$
The numerator grows faster, so the function grows without bound. No horizontal asymptote.
This technique works because when $x$ is enormous, only the highest-degree terms in the numerator and denominator matter. All other terms become relatively insignificant.
Vertical Asymptotes and Infinite Limits
While limits at infinity ask what happens as $x$ goes to infinity, infinite limits ask what happens when the function values go to infinity. These are different questions with different answers.
A vertical asymptote occurs at $x = a$ when the function “blows up” (goes to positive or negative infinity) as $x$ approaches $a$. This typically happens when the denominator of a fraction approaches zero while the numerator does not.
For example, $f(x) = \frac{1}{x-3}$ has a vertical asymptote at $x = 3$. As $x$ approaches 3, the denominator gets tiny, making the fraction huge:
$$\lim_{x \to 3^+} \frac{1}{x-3} = +\infty \quad \text{and} \quad \lim_{x \to 3^-} \frac{1}{x-3} = -\infty$$
The superscripts $3^+$ and $3^-$ indicate approaching 3 from the right (values slightly greater than 3) and from the left (values slightly less than 3), respectively.
Important distinction:
- $\lim_{x \to \infty} f(x) = L$ means $x$ goes to infinity, $f(x)$ approaches some finite value $L$
- $\lim_{x \to a} f(x) = \infty$ means $x$ approaches some finite value $a$, $f(x)$ goes to infinity
End Behavior
The end behavior of a function describes how the function behaves as $x$ goes to positive infinity and negative infinity. For any function, we might ask:
- Does $f(x)$ approach a finite limit as $x \to \infty$?
- Does $f(x)$ approach a finite limit as $x \to -\infty$?
- If not, does $f(x)$ go to $+\infty$ or $-\infty$?
For polynomials, end behavior is determined entirely by the leading term. For $f(x) = -2x^3 + 100x^2 - 5$:
- The leading term $-2x^3$ dominates for large $|x|$
- As $x \to \infty$: $-2x^3 \to -\infty$, so $f(x) \to -\infty$
- As $x \to -\infty$: $-2(-\infty)^3 = -2(-\infty) = +\infty$, so $f(x) \to +\infty$
Notation and Terminology
| Term | Meaning | Example |
|---|---|---|
| $\lim_{x \to \infty} f(x) = L$ | $f(x)$ approaches $L$ as $x$ grows without bound | $\lim_{x \to \infty} \frac{1}{x} = 0$ |
| $\lim_{x \to -\infty} f(x) = L$ | $f(x)$ approaches $L$ as $x$ decreases without bound | $\lim_{x \to -\infty} \frac{1}{x} = 0$ |
| Horizontal asymptote | Line $y = L$ that function approaches as $x \to \pm\infty$ | $y = 2$ for $f(x) = \frac{2x}{x+1}$ |
| Vertical asymptote | Line $x = a$ where function values approach $\pm\infty$ | $x = 0$ for $f(x) = \frac{1}{x}$ |
| $\lim_{x \to a} f(x) = \infty$ | $f(x)$ grows without bound as $x$ approaches $a$ | $\lim_{x \to 0^+} \frac{1}{x} = \infty$ |
| End behavior | How function behaves as $x \to \pm\infty$ | $x^2 \to \infty$ as $x \to \pm\infty$ |
| Leading term | Highest-degree term in a polynomial | $3x^4$ in $3x^4 - 2x + 1$ |
| Leading coefficient | Coefficient of the leading term | $3$ in $3x^4 - 2x + 1$ |
| Degree | Highest exponent in a polynomial | Degree of $3x^4 - 2x + 1$ is 4 |
Examples
Find $\lim_{x \to \infty} \frac{3x}{x + 1}$.
Solution:
This is a rational function where both numerator and denominator have degree 1 (same degree). We can use the degree comparison technique: the limit equals the ratio of leading coefficients.
Leading coefficient of numerator: 3 Leading coefficient of denominator: 1
$$\lim_{x \to \infty} \frac{3x}{x + 1} = \frac{3}{1} = 3$$
Alternative approach: Divide numerator and denominator by $x$:
$$\lim_{x \to \infty} \frac{3x}{x + 1} = \lim_{x \to \infty} \frac{3}{1 + \frac{1}{x}}$$
As $x \to \infty$, $\frac{1}{x} \to 0$, so:
$$= \frac{3}{1 + 0} = 3$$
The horizontal asymptote is $y = 3$.
Identify the horizontal asymptote of $f(x) = \frac{2}{x - 5}$.
Solution:
The numerator has degree 0 (it is just the constant 2), and the denominator has degree 1. Since the degree of the numerator is less than the degree of the denominator, the limit at infinity is 0:
$$\lim_{x \to \infty} \frac{2}{x - 5} = 0$$
To verify: when $x$ is very large (say, $x = 1,000,000$), we get $\frac{2}{999,995} \approx 0.000002$, which is very close to zero.
The horizontal asymptote is $y = 0$ (the $x$-axis).
Bonus: This function also has a vertical asymptote at $x = 5$, where the denominator equals zero.
Find $\lim_{x \to \infty} \frac{2x^2 + 3}{5x^2 - x}$.
Solution:
Both numerator and denominator have degree 2. The limit equals the ratio of leading coefficients:
$$\lim_{x \to \infty} \frac{2x^2 + 3}{5x^2 - x} = \frac{2}{5}$$
Let us verify by the division technique: Divide every term by $x^2$ (the highest power present):
$$\lim_{x \to \infty} \frac{2x^2 + 3}{5x^2 - x} = \lim_{x \to \infty} \frac{2 + \frac{3}{x^2}}{5 - \frac{1}{x}}$$
As $x \to \infty$:
- $\frac{3}{x^2} \to 0$
- $\frac{1}{x} \to 0$
So:
$$= \frac{2 + 0}{5 - 0} = \frac{2}{5}$$
The horizontal asymptote is $y = \frac{2}{5}$.
Find all asymptotes of $f(x) = \frac{x^2 - 4}{x - 2}$.
Solution:
Step 1: Look for vertical asymptotes.
Vertical asymptotes occur where the denominator equals zero (and the numerator does not). Setting $x - 2 = 0$ gives $x = 2$.
But wait. Let us factor the numerator: $x^2 - 4 = (x-2)(x+2)$.
$$f(x) = \frac{(x-2)(x+2)}{x-2}$$
The factor $(x-2)$ cancels:
$$f(x) = x + 2 \quad \text{for } x \neq 2$$
Since the problematic factor cancels, there is no vertical asymptote at $x = 2$. Instead, there is a “hole” in the graph at $x = 2$.
Step 2: Look for horizontal asymptotes.
After simplification, $f(x) = x + 2$ is a linear function (a polynomial of degree 1). Linear functions grow without bound as $x \to \pm\infty$:
$$\lim_{x \to \infty} (x + 2) = \infty$$
So there is no horizontal asymptote.
Step 3: Check for slant (oblique) asymptotes.
Since the original function simplifies to a linear function, the line $y = x + 2$ is actually what the function equals everywhere except at the hole. The graph is just the line $y = x + 2$ with a single point missing at $x = 2$.
Summary: No vertical asymptotes, no horizontal asymptotes, just a hole at $(2, 4)$.
Analyze $\lim_{x \to \infty} \frac{3x^3 - 2x}{x^2 + 1}$.
Solution:
The numerator has degree 3, and the denominator has degree 2. Since the numerator has the higher degree, the function grows without bound. There is no horizontal asymptote.
But we can be more precise about how fast it grows. Divide numerator and denominator by $x^2$:
$$\frac{3x^3 - 2x}{x^2 + 1} = \frac{3x - \frac{2}{x}}{1 + \frac{1}{x^2}}$$
As $x \to \infty$, the terms $\frac{2}{x}$ and $\frac{1}{x^2}$ vanish, leaving:
$$\approx \frac{3x}{1} = 3x$$
So the function behaves like $3x$ for large $x$. This means there is a slant (oblique) asymptote at $y = 3x$.
To find the exact slant asymptote, we can perform polynomial long division:
$$\frac{3x^3 - 2x}{x^2 + 1} = 3x + \frac{-5x}{x^2 + 1}$$
Wait, let us redo that more carefully. Dividing $3x^3 - 2x$ by $x^2 + 1$:
- $3x^3 \div x^2 = 3x$
- $3x \cdot (x^2 + 1) = 3x^3 + 3x$
- Subtract: $(3x^3 - 2x) - (3x^3 + 3x) = -5x$
So: $\frac{3x^3 - 2x}{x^2 + 1} = 3x + \frac{-5x}{x^2 + 1}$
As $x \to \infty$, $\frac{-5x}{x^2 + 1} \to 0$ (degree 1 over degree 2), so the function approaches the line $y = 3x$.
Summary:
- $\lim_{x \to \infty} f(x) = \infty$ (no horizontal asymptote)
- Slant asymptote: $y = 3x$
- No vertical asymptotes (denominator $x^2 + 1$ is never zero)
Find all asymptotes and describe the end behavior of $f(x) = \frac{x}{\sqrt{x^2 + 1}}$.
Solution:
Horizontal asymptotes:
For $x \to +\infty$:
We can factor $x^2$ out of the square root, but we must be careful. For $x > 0$, $\sqrt{x^2} = |x| = x$.
$$\frac{x}{\sqrt{x^2 + 1}} = \frac{x}{\sqrt{x^2(1 + \frac{1}{x^2})}} = \frac{x}{|x|\sqrt{1 + \frac{1}{x^2}}}$$
For $x > 0$, $|x| = x$:
$$= \frac{x}{x\sqrt{1 + \frac{1}{x^2}}} = \frac{1}{\sqrt{1 + \frac{1}{x^2}}}$$
As $x \to +\infty$, $\frac{1}{x^2} \to 0$:
$$\lim_{x \to +\infty} f(x) = \frac{1}{\sqrt{1 + 0}} = 1$$
For $x \to -\infty$:
For $x < 0$, $|x| = -x$:
$$\frac{x}{|x|\sqrt{1 + \frac{1}{x^2}}} = \frac{x}{-x\sqrt{1 + \frac{1}{x^2}}} = \frac{-1}{\sqrt{1 + \frac{1}{x^2}}}$$
As $x \to -\infty$:
$$\lim_{x \to -\infty} f(x) = \frac{-1}{\sqrt{1}} = -1$$
Summary:
- Horizontal asymptote $y = 1$ as $x \to +\infty$
- Horizontal asymptote $y = -1$ as $x \to -\infty$
- No vertical asymptotes (denominator is always at least 1)
This function has two different horizontal asymptotes, one in each direction.
Key Properties and Rules
Fundamental Limits at Infinity
$$\lim_{x \to \infty} \frac{1}{x^n} = 0 \quad \text{for any } n > 0$$
$$\lim_{x \to \infty} c = c \quad \text{for any constant } c$$
$$\lim_{x \to \infty} x^n = \infty \quad \text{for any } n > 0$$
Limit Laws Still Apply
If $\lim_{x \to \infty} f(x) = L$ and $\lim_{x \to \infty} g(x) = M$, then:
$$\lim_{x \to \infty} [f(x) + g(x)] = L + M$$
$$\lim_{x \to \infty} [f(x) \cdot g(x)] = L \cdot M$$
$$\lim_{x \to \infty} \frac{f(x)}{g(x)} = \frac{L}{M} \quad \text{(if } M \neq 0\text{)}$$
Rational Function Rules
For $\frac{a_n x^n + \ldots + a_0}{b_m x^m + \ldots + b_0}$:
- If $n < m$: limit is $0$
- If $n = m$: limit is $\frac{a_n}{b_m}$
- If $n > m$: limit is $\pm\infty$ (sign depends on leading coefficients)
The Division Technique
To evaluate a limit at infinity for a rational function, divide every term by the highest power of $x$ in the denominator. This converts terms with $x$ in the denominator into forms that approach zero.
Vertical Asymptotes
To find vertical asymptotes of a rational function:
- Factor numerator and denominator completely
- Cancel any common factors (these create holes, not asymptotes)
- Set the remaining denominator equal to zero
- Each solution is a vertical asymptote
Infinite Limits: Determining the Sign
When $\lim_{x \to a} f(x) = \pm\infty$:
- Analyze the sign of the numerator near $x = a$
- Analyze the sign of the denominator near $x = a$ (from left and right)
- The sign of the limit is the sign of their quotient
Real-World Applications
Terminal Velocity
When you drop an object through air, gravity accelerates it downward, but air resistance pushes back. Air resistance increases with speed, so eventually the two forces balance and the object reaches a constant terminal velocity.
The velocity function might look like:
$$v(t) = v_{terminal}(1 - e^{-kt})$$
As $t \to \infty$, $e^{-kt} \to 0$, so $v(t) \to v_{terminal}$. The horizontal asymptote represents the maximum speed the object can achieve.
For a skydiver, $v_{terminal} \approx 55$ m/s (about 120 mph). No matter how long they fall, they will not exceed this speed (until they deploy their parachute, which changes the equation entirely).
Drug Concentration in the Bloodstream
When medication is administered intravenously at a constant rate, the drug concentration in the blood eventually stabilizes. The body eliminates the drug at a rate proportional to its concentration, leading to a model like:
$$C(t) = C_{max}(1 - e^{-rt})$$
The horizontal asymptote $C_{max}$ represents the equilibrium concentration, where the rate of drug entering equals the rate being eliminated. Understanding this limit helps doctors determine proper dosing.
Population Growth and Carrying Capacity
In ecology, the logistic growth model accounts for limited resources:
$$P(t) = \frac{K}{1 + Ae^{-rt}}$$
where $K$ is the carrying capacity, the maximum sustainable population. As $t \to \infty$:
$$\lim_{t \to \infty} P(t) = \frac{K}{1 + 0} = K$$
The carrying capacity is a horizontal asymptote. A population might grow rapidly at first, but it levels off as resources become scarce. This model applies to bacteria in a petri dish, deer in a forest, or even the spread of ideas through a population.
Economics: Market Saturation
When a new product enters the market, sales often follow an S-curve. Early adopters buy quickly, the middle majority follows, but eventually nearly everyone who wants the product has it. Total sales approach a limit (market saturation), which is a horizontal asymptote.
Similarly, the price of a commodity in a competitive market approaches an equilibrium price as supply and demand adjust. The equilibrium is a horizontal asymptote for the price function over time.
Learning Curves
Your performance at a new skill often follows a pattern: rapid improvement at first, then slower gains as you approach your maximum capability. A model might be:
$$\text{Skill}(t) = M(1 - e^{-kt})$$
The horizontal asymptote $M$ represents your maximum potential in that skill. This explains why the first few hours of practice produce dramatic results, while later hours yield diminishing returns.
Self-Test Problems
Problem 1: Find $\lim_{x \to \infty} \frac{5x^2 - 3x + 1}{2x^2 + 4x - 7}$.
Show Answer
Both numerator and denominator have degree 2. The limit is the ratio of leading coefficients:
$$\lim_{x \to \infty} \frac{5x^2 - 3x + 1}{2x^2 + 4x - 7} = \frac{5}{2}$$
The horizontal asymptote is $y = \frac{5}{2}$.
Problem 2: Identify all asymptotes of $f(x) = \frac{3}{(x-1)(x+2)}$.
Show Answer
Vertical asymptotes: Set denominator equal to zero: $(x-1)(x+2) = 0$ gives $x = 1$ and $x = -2$.
Vertical asymptotes at $x = 1$ and $x = -2$.
Horizontal asymptote: The numerator has degree 0, the denominator has degree 2. Since degree of numerator < degree of denominator:
$$\lim_{x \to \infty} f(x) = 0$$
Horizontal asymptote at $y = 0$.
Problem 3: Evaluate $\lim_{x \to \infty} \frac{4x^3 + x}{2x^3 - 5}$.
Show Answer
Both have degree 3, so the limit is the ratio of leading coefficients:
$$\lim_{x \to \infty} \frac{4x^3 + x}{2x^3 - 5} = \frac{4}{2} = 2$$
Problem 4: What is $\lim_{x \to \infty} \frac{x^2 + 1}{x^3 - x}$?
Show Answer
The numerator has degree 2, the denominator has degree 3. Since degree of numerator < degree of denominator:
$$\lim_{x \to \infty} \frac{x^2 + 1}{x^3 - x} = 0$$
You can verify by dividing by $x^3$:
$$\frac{\frac{1}{x} + \frac{1}{x^3}}{1 - \frac{1}{x^2}} \to \frac{0 + 0}{1 - 0} = 0$$
Problem 5: Find $\lim_{x \to 2^+} \frac{x+1}{x-2}$.
Show Answer
As $x \to 2^+$ (approaching 2 from the right):
- Numerator: $x + 1 \to 3$ (positive)
- Denominator: $x - 2 \to 0^+$ (small positive)
A positive number divided by a small positive number gives a large positive number:
$$\lim_{x \to 2^+} \frac{x+1}{x-2} = +\infty$$
Problem 6: Find all asymptotes of $g(x) = \frac{2x^2 - 8}{x - 2}$.
Show Answer
First, factor the numerator: $2x^2 - 8 = 2(x^2 - 4) = 2(x-2)(x+2)$.
$$g(x) = \frac{2(x-2)(x+2)}{x-2} = 2(x+2) = 2x + 4 \quad \text{for } x \neq 2$$
Since the $(x-2)$ factor cancels, there is no vertical asymptote at $x = 2$, just a hole at the point $(2, 8)$.
After simplification, $g(x) = 2x + 4$ is a linear function, which has no horizontal asymptote. However, the line $y = 2x + 4$ is exactly what the function equals (except at the hole).
Summary: No vertical asymptotes, no horizontal asymptotes. The graph is the line $y = 2x + 4$ with a hole at $(2, 8)$.
Problem 7: Evaluate $\lim_{x \to -\infty} \frac{3x}{|x| + 1}$.
Show Answer
For $x < 0$, $|x| = -x$, so:
$$\frac{3x}{|x| + 1} = \frac{3x}{-x + 1}$$
Divide numerator and denominator by $x$ (noting $x < 0$):
$$= \frac{3}{-1 + \frac{1}{x}}$$
As $x \to -\infty$, $\frac{1}{x} \to 0$:
$$\lim_{x \to -\infty} \frac{3x}{|x| + 1} = \frac{3}{-1 + 0} = -3$$
Problem 8: A population follows the model $P(t) = \frac{1000}{1 + 49e^{-0.3t}}$. Find the carrying capacity (the long-term population limit).
Show Answer
As $t \to \infty$, $e^{-0.3t} \to 0$:
$$\lim_{t \to \infty} P(t) = \frac{1000}{1 + 49 \cdot 0} = \frac{1000}{1} = 1000$$
The carrying capacity is 1000. No matter how much time passes, the population will approach but not exceed 1000.
Summary
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Limits at infinity describe function behavior as $x$ grows without bound. We write $\lim_{x \to \infty} f(x) = L$ to mean $f(x)$ approaches $L$ as $x$ increases forever.
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Horizontal asymptotes are horizontal lines $y = L$ that a function approaches as $x \to \pm\infty$. A function can have zero, one, or two horizontal asymptotes.
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Vertical asymptotes occur at values $x = a$ where the function blows up to $\pm\infty$. These happen when the denominator approaches zero while the numerator does not.
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For rational functions, compare degrees of numerator and denominator:
- Degree of top < degree of bottom: limit is 0
- Degree of top = degree of bottom: limit is ratio of leading coefficients
- Degree of top > degree of bottom: no horizontal asymptote (function grows without bound)
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The division technique (dividing all terms by the highest power in the denominator) is a reliable way to evaluate limits at infinity algebraically.
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Growth rate hierarchy determines which terms dominate: higher-degree terms overwhelm lower-degree terms when $x$ is large.
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End behavior describes how a function behaves in both directions ($x \to +\infty$ and $x \to -\infty$). These can be different.
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Limits at infinity have wide applications: terminal velocity, drug concentrations, population limits, market saturation, and learning curves are all described by horizontal asymptotes.
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Do not confuse $\lim_{x \to \infty} f(x) = L$ (the variable goes to infinity, the function approaches a finite value) with $\lim_{x \to a} f(x) = \infty$ (the variable approaches a finite value, the function goes to infinity).