So far, we have studied the behavior of functions as $x$ approaches a specific number, $a$. Now, we shift our perspective to understand the end behavior of a function. What happens to the values of $f(x)$ as $x$ becomes incredibly large in the positive direction ($x \to \infty$) or incredibly large in the negative direction ($x \to -\infty$)? This concept is crucial for understanding phenomena with long-term trends, and it gives rise to the idea of horizontal asymptotes. ↔️
Let's formalize the idea of a function leveling off at a certain height as $x$ goes to infinity.
Let $f$ be a function defined on some interval $(a, \infty)$. Then $$ \lim_{x \to \infty} f(x) = L $$ means that the values of $f(x)$ can be made arbitrarily close to $L$ by taking $x$ sufficiently large.
Similarly, if $f$ is defined on $(-\infty, a)$, then $$ \lim_{x \to -\infty} f(x) = L $$ means that the values of $f(x)$ can be made arbitrarily close to $L$ by taking $x$ sufficiently large negatively.
The line $y=L$ is called a horizontal asymptote of the curve $y=f(x)$ if either $$ \lim_{x \to \infty} f(x) = L \quad \text{or} \quad \lim_{x \to -\infty} f(x) = L $$
Consider the function $f(x) = \frac{1}{x}$. As $x$ gets very large (e.g., 100, 1000, 1,000,000), the value of $\frac{1}{x}$ gets very close to 0. Similarly, as $x$ becomes a large negative number (e.g., -100, -1000), $\frac{1}{x}$ also gets very close to 0.
$$ \lim_{x \to \infty} \frac{1}{x} = 0 \quad \text{and} \quad \lim_{x \to -\infty} \frac{1}{x} = 0 $$
Therefore, the line $y=0$ is a horizontal asymptote for $f(x) = \frac{1}{x}$.
Try it yourself: In the Desmos graph below, type the function $f(x)=1/x$ on the first line and the asymptote $y=0$ on the second line to visualize this relationship.
The previous example illustrates a powerful theorem that is the key to finding limits at infinity for any rational function.
If $r > 0$ is a rational number, then $$ \lim_{x \to \infty} \frac{1}{x^r} = 0 $$
If $r > 0$ is a rational number such that $x^r$ is defined for all $x$, then $$ \lim_{x \to -\infty} \frac{1}{x^r} = 0 $$
For instance, for $r=2$, we have $\lim_{x \to \infty} \frac{1}{x^2} = 0$. The second part of the theorem is important for limits to negative infinity. The phrase "such that $x^r$ is defined for all $x$" means we must be able to evaluate the function for negative numbers.
This theorem is our fundamental tool.
How do we handle more complex functions, like rational functions, which are ratios of polynomials? For instance, how would we evaluate $\lim_{x \to \infty} \frac{3x^2 - x - 2}{5x^2 + 4x + 1}$? We can't just substitute $\infty$ and get $\frac{\infty}{\infty}$, which is an indeterminate form.
It's important to note that all the Limit Laws we established in a previous lecture (for Sum, Difference, Product, Quotient, etc.) are also valid for limits as $x \to \infty$ and $x \to -\infty$. This is what allows us to validly evaluate the limit of the numerator and the denominator separately in our upcoming examples, after we perform our algebraic trick.
To solve this, we use a clever algebraic trick. The main strategy is to divide the numerator and denominator by the highest power of $x$ that appears in the denominator. This transforms the problem into several terms that go to zero.
Find the limit: $\lim_{x \to \infty} \frac{3x^2 - x - 2}{5x^2 + 4x + 1}$.
The highest power of $x$ in the denominator is $x^2$. We divide every term by $x^2$.
$$ \lim_{x \to \infty} \frac{\frac{3x^2}{x^2} - \frac{x}{x^2} - \frac{2}{x^2}}{\frac{5x^2}{x^2} + \frac{4x}{x^2} + \frac{1}{x^2}} = \lim_{x \to \infty} \frac{3 - \frac{1}{x} - \frac{2}{x^2}}{5 + \frac{4}{x} + \frac{1}{x^2}} $$Now, we use the theorem. As $x \to \infty$, all terms of the form $\frac{c}{x^r}$ approach 0.
$$ = \frac{3 - 0 - 0}{5 + 0 + 0} = \frac{3}{5} $$Thus, the line $y=3/5$ is a horizontal asymptote.
In the Desmos graph, confirm this result by plotting the function and its horizontal asymptote, $y=3/5$.
Find the limit: $\lim_{x \to \infty} \frac{x^3-2x}{x^2+1}$.
For a rational function $f(x) = \frac{P(x)}{Q(x)}$, where the degree of $P(x)$ is $n$ and the degree of $Q(x)$ is $m$:
When dealing with radicals, the same principle applies, but we must be careful, especially when $x \to -\infty$. Remember that for $x < 0$, we have $\sqrt{x^2} = |x| = -x$.
Find the horizontal asymptote of $f(x) = \frac{\sqrt{9x^2+2}}{4x+5}$ for $x < 0$.
We need to evaluate $\lim_{x \to -\infty} \frac{\sqrt{9x^2+2}}{4x+5}$. The highest power of $x$ in the denominator is $x^1$. We divide the numerator and denominator by $x$.
For the numerator, since $x$ is negative, $x = -\sqrt{x^2}$. So, dividing by $x$ is the same as dividing by $-\sqrt{x^2}$.
$$ \lim_{x \to -\infty} \frac{\frac{\sqrt{9x^2+2}}{-\sqrt{x^2}}}{\frac{4x+5}{x}} = \lim_{x \to -\infty} \frac{-\sqrt{\frac{9x^2+2}{x^2}}}{\frac{4x}{x}+\frac{5}{x}} $$ $$ = \lim_{x \to -\infty} \frac{-\sqrt{9+\frac{2}{x^2}}}{4+\frac{5}{x}} $$As $x \to -\infty$, the terms $\frac{2}{x^2}$ and $\frac{5}{x}$ go to 0.
$$ = \frac{-\sqrt{9+0}}{4+0} = -\frac{3}{4} $$So, the line $y = -3/4$ is a horizontal asymptote as $x \to -\infty$.
Sometimes, a function doesn't approach a finite number but grows without bound as $x$ approaches infinity. We call these infinite limits at infinity.
Evaluate $\lim_{x \to \infty} (x^3 - x)$.
We can't use limit laws on $\infty - \infty$ (an indeterminate form). Instead, we factor out the highest power of $x$.
$$ \lim_{x \to \infty} x^3(1 - \frac{1}{x^2}) $$As $x \to \infty$, $x^3 \to \infty$ and $(1 - \frac{1}{x^2}) \to (1-0) = 1$. A very large number multiplied by a number close to 1 is a very large number.
$$ \lim_{x \to \infty} (x^3 - x) = \infty $$Problem: Find the limit $\lim_{x \to \infty} \arctan(e^{-x})$.
Find the limit: $\lim_{x \to - \infty} \frac{x^2+1}{2x^3 - 3x + 4}$.
Find the horizontal asymptotes of the function $f(x) = \frac{5x-2}{\sqrt{x^2+1}}$. (Hint: You need to check both $x \to \infty$ and $x \to -\infty$).
Evaluate the limit: $\lim_{x \to \infty} \sin(1/x)$.
Today we extended our understanding of limits to describe the end behavior of functions. By analyzing what happens as $x$ approaches $\infty$ and $-\infty$, we can identify horizontal asymptotes, which provide a powerful summary of a function's long-term trend. The key technique of dividing by the highest power of $x$ in the denominator is a fundamental skill that you will use throughout calculus. This understanding of end behavior, combined with our knowledge of behavior near a point, gives us a much more complete picture of a function's graph. Next, we will finally introduce the concept of the derivative! 📈