2.2 – Limits Involving Infinity

Infinity

Infinity ($\infty$) does not represent a real number. When we evaluate the limit of a function $f(x)$ as $x$ approaches infinity, we observe the behaviour of the function to the right extremes on the number line.

Examples

Finite Limits as $x\to \infty$

Observing a table of values for the function $f(x)=\frac{1}{x}$:

$x$	$f(x)$
$\to -\infty$	0${}^{-}$
-1,000	-0.001
-100	-0.01
-10	-0.1
-1	-1
0	undefined
1	1
10	0.1
100	0.01
1,000	0.001
$\to +\infty$	0${}^{+}$

We find that $f(x)$ approaches zero. Therefore, the line $y=0$ is a horizontal asymptote of the function.

Knowing the asymptomatic behaviour of the function, we can write the limits at either infinity as

$$\underset{x\to -\infty }{\lim }\frac{1}{x}=0\text{ and }\underset{x\to +\infty }{\lim }\frac{1}{x}=0$$

Horizontal Asymptotes

Functions with horizontal asymptotes have finite limits as $x\to \pm \infty$.

Infinite Limits as $x\to a$

Observing a table of the same function as $x$ approaches zero:

$x$	$f(x)$
-1	-1
-0.1	-10
-0.01	-100
-0.001	-1,000
0${}^{-}$	$\to -\infty$
0	undefined
0${}^{+}$	$\to +\infty$
0.001	1,000
0.01	100
0.1	10
1	1

We find that $f(x)$ grows unbounded near zero. Therefore, the line $x=0$ is a vertical asymptote of the function.

Vertical Asymptotes

Functions have horizontal asymptotes at the line $x=a$ if either

$$\underset{x\to a^{-}}{\lim }f(x)=\pm \infty \text{ or }\underset{x\to a^{+}}{\lim }f(x)=\pm \infty$$

Polynomials

When finding the limits of polynomials at infinity, it is only necessary to look at the highest-degree term. One way to do this is by dividing each term by $x$ to the highest degree.

$$\underset{x\to \infty }{\lim }\frac{3x^3+2x^2-5}{x^2-4x^3}$$ $$\underset{x\to \infty }{\lim }\frac{\frac{3x^3}{x^3}+\frac{2x^2}{x^3}-\frac{5}{x^3}}{\frac{x^2}{x^3}-\frac{4x^3}{x^3}}$$ $$\underset{x\to \infty }{\lim }\frac{3+\frac{2}{x}-\frac{5}{x^3}}{\frac{1}{x}-4}$$ $$\underset{x\to \infty }{\lim }\frac{3}{-4}=-\frac{3}{4}$$

Examples

$$\underset{x\to \infty }{\lim }\frac{x^2+2x-5}{x-3}\Rightarrow \underset{x\to \infty }{\lim }{x}^2=\infty$$ $$\underset{x\to -\infty }{\lim }\frac{2x^3-4x+6}{x^2-4}\Rightarrow \underset{x\to -\infty }{\lim }2x=-\infty$$ $$\underset{x\to \infty }{\lim }\frac{3x^3+2x^2-5}{x^2-4x^3}\Rightarrow \underset{x\to \infty }{\lim }-\frac{3}{4}=-\frac{3}{4}$$

End-Behaviour Models