Infinite limits and limits at infinity
This page covers two ideas that both use the symbol $\infty$ (infinity):
- Infinite limits: $f(x)$ increases or decreases without bound as $x$ approaches a number $a$.
- Limits at infinity: $x$ increases or decreases without bound, and we ask what $f(x)$ approaches.
Infinite limits
Look at the function $f(x)=\dfrac{1}{(x-1)^2}$ near $x=1$.
As $x$ approaches $1$, the denominator $(x-1)^2$ approaches $0$. It is always positive. So $f(x)$ becomes a very large positive number. For example, $f(1.01)=\dfrac{1}{0.0001}=10000$.
We say that $f(x)$ increases without bound, and we write
$\displaystyle\lim_{x\to 1}\dfrac{1}{(x-1)^2}=\infty$.
Important: $\infty$ is not a number. So this limit does not exist. Writing $=\infty$ tells us how the limit fails to exist: the values increase without bound.
In the same way, if $f(x)$ decreases without bound as $x\to a$, we write
$\displaystyle\lim_{x\to a}f(x)=-\infty$.
One-sided infinite limits
Sometimes a function goes up on one side and down on the other side. Look at $f(x)=\dfrac{1}{x-2}$ near $x=2$.
- If $x$ is a little more than $2$, then $x-2$ is a small positive number. So $\dfrac{1}{x-2}$ is a large positive number.
- If $x$ is a little less than $2$, then $x-2$ is a small negative number. So $\dfrac{1}{x-2}$ is a large negative number.
So,
$\displaystyle\lim_{x\to 2^+}\dfrac{1}{x-2}=\infty$ and $\displaystyle\lim_{x\to 2^-}\dfrac{1}{x-2}=-\infty$.
Vertical asymptotes
In both graphs above, the graph gets closer and closer to a vertical line. This line is called a vertical asymptote.
Definition: The line $x=a$ is a vertical asymptote of $f$ if at least one of the one-sided limits at $a$ is $\infty$ or $-\infty$.
For a rational function, vertical asymptotes can occur where the denominator is $0$. But first cancel any common factors. A factor that cancels gives a hole in the graph, not an asymptote.
Finding infinite limits
When direct substitution gives a nonzero number over $0$, the limit is $\infty$ or $-\infty$ (or one of each, from the two sides). To decide which, look at the signs:
- Find the sign of the numerator near $a$.
- Find the sign of the denominator just to the left and just to the right of $a$.
- If the numerator and the denominator have the same sign, the limit is $\infty$. If they have different signs, the limit is $-\infty$.
Example 1: Find $\displaystyle\lim_{x\to 3^+}\dfrac{2}{x-3}$ and $\displaystyle\lim_{x\to 3^-}\dfrac{2}{x-3}$.
Solution:
The numerator is $2$, which is positive.
From the right ($x>3$), the denominator $x-3$ is a small positive number. Positive over positive is positive. So,
$\displaystyle\lim_{x\to 3^+}\dfrac{2}{x-3}=\infty$.
From the left ($x<3$), the denominator $x-3$ is a small negative number. Positive over negative is negative. So,
$\displaystyle\lim_{x\to 3^-}\dfrac{2}{x-3}=-\infty$.
Example 2: Find $\displaystyle\lim_{x\to -1}\dfrac{x+4}{(x+1)^2}$.
Solution:
As $x\to -1$, the numerator approaches $-1+4=3$, which is positive.
The denominator $(x+1)^2$ approaches $0$. It is a square, so it is positive on both sides of $-1$.
Positive over a small positive number gives a large positive number on both sides. So,
$\displaystyle\lim_{x\to -1}\dfrac{x+4}{(x+1)^2}=\infty$.
Example 3: Find the vertical asymptotes of $f(x)=\dfrac{x-2}{x^2-4}$.
Solution:
The denominator is $0$ at $x=2$ and at ${x=-2}$. First, factor the denominator:
$\begin{align*}&f(x)\\&=\dfrac{x-2}{x^2-4}\\&=\dfrac{x-2}{(x-2)(x+2)}\end{align*}$
Cancel the common factor $x-2$:
$\begin{align*}&f(x)=\dfrac{1}{x+2},\ \text{for }x\neq 2\end{align*}$
At $x=-2$, the denominator $x+2$ is $0$ and the numerator is $1$. So $f$ has infinite one-sided limits at $-2$. The line $x=-2$ is a vertical asymptote.
At $x=2$, the factor $x-2$ cancels. Here the limit is a number:
$\displaystyle\lim_{x\to 2}f(x)=\dfrac{1}{2+2}=\dfrac{1}{4}$.
So the graph has a hole at $x=2$, not an asymptote. The only vertical asymptote is $x=-2$.
Limits at infinity
Now we ask a different question: what happens to $f(x)$ when $x$ gets larger and larger?
We write $x\to\infty$ to mean that $x$ increases without bound. We write $x\to -\infty$ to mean that $x$ decreases without bound (becomes a larger and larger negative number).
If $f(x)$ approaches a number $L$ as $x\to\infty$, we write
$\displaystyle\lim_{x\to\infty}f(x)=L$.
The most important example is $\dfrac{1}{x}$. As $x$ gets larger, $\dfrac{1}{x}$ gets closer to $0$:
| $x$ | $\dfrac{1}{x}$ |
|---|---|
| 10 | 0.1 |
| 100 | 0.01 |
| 1000 | 0.001 |
| 10000 | 0.0001 |
So,
$\displaystyle\lim_{x\to\infty}\dfrac{1}{x}=0$ and $\displaystyle\lim_{x\to -\infty}\dfrac{1}{x}=0$.
In the same way, for any positive integer $n$,
$\displaystyle\lim_{x\to\infty}\dfrac{1}{x^n}=0$ and $\displaystyle\lim_{x\to -\infty}\dfrac{1}{x^n}=0$.
In words: a fixed number divided by a larger and larger number approaches $0$.
Horizontal asymptotes
Definition: The line $y=L$ is a horizontal asymptote of $f$ if
$\displaystyle\lim_{x\to\infty}f(x)=L$ or $\displaystyle\lim_{x\to -\infty}f(x)=L$.
The graph levels off and gets closer and closer to the line $y=L$.
Limits at infinity of rational functions
To find the limit of a rational function as $x\to\infty$ or $x\to -\infty$:
- Find the highest power of $x$ in the denominator.
- Divide every term in the numerator and the denominator by that power.
- Use $\dfrac{1}{x^n}\to 0$ for each term that has $x$ in its denominator.
Example 4: Find $\displaystyle\lim_{x\to\infty}\dfrac{3x^2-x+2}{5x^2+4x-1}$.
Solution:
The highest power of $x$ in the denominator is $x^2$. Divide every term by $x^2$:
$\begin{align*}&\lim_{x\to\infty}\dfrac{3x^2-x+2}{5x^2+4x-1}\\&=\lim_{x\to\infty}\dfrac{3-\dfrac{1}{x}+\dfrac{2}{x^2}}{5+\dfrac{4}{x}-\dfrac{1}{x^2}}\\&=\dfrac{3-0+0}{5+0-0}\\&=\dfrac{3}{5}\end{align*}$
So $y=\dfrac{3}{5}$ is a horizontal asymptote.
Example 5: Find $\displaystyle\lim_{x\to\infty}\dfrac{2x+7}{x^2+1}$.
Solution:
The highest power of $x$ in the denominator is $x^2$. Divide every term by $x^2$:
$\begin{align*}&\lim_{x\to\infty}\dfrac{2x+7}{x^2+1}\\&=\lim_{x\to\infty}\dfrac{\dfrac{2}{x}+\dfrac{7}{x^2}}{1+\dfrac{1}{x^2}}\\&=\dfrac{0+0}{1+0}\\&=0\end{align*}$
So $y=0$ (the $x$-axis) is a horizontal asymptote.
Example 6: Find $\displaystyle\lim_{x\to\infty}\dfrac{x^2+1}{x-3}$.
Solution:
The highest power of $x$ in the denominator is $x$. Divide every term by $x$:
$\begin{align*}&\lim_{x\to\infty}\dfrac{x^2+1}{x-3}\\&=\lim_{x\to\infty}\dfrac{x+\dfrac{1}{x}}{1-\dfrac{3}{x}}\end{align*}$
The denominator approaches $1$. But the numerator $x+\dfrac{1}{x}$ increases without bound. So,
$\displaystyle\lim_{x\to\infty}\dfrac{x^2+1}{x-3}=\infty$.
This function has no horizontal asymptote on the right.
Example 7: Find $\displaystyle\lim_{x\to -\infty}\dfrac{4x^3+x}{2x^3-7}$.
Solution:
The method also works as $x\to -\infty$. The highest power of $x$ in the denominator is $x^3$. Divide every term by $x^3$:
$\begin{align*}&\lim_{x\to -\infty}\dfrac{4x^3+x}{2x^3-7}\\&=\lim_{x\to -\infty}\dfrac{4+\dfrac{1}{x^2}}{2-\dfrac{7}{x^3}}\\&=\dfrac{4+0}{2-0}\\&=2\end{align*}$
A shortcut for rational functions
The degree of a polynomial is its highest power of $x$. The leading coefficient is the number in front of that highest power. The examples above show a pattern. Compare the degree of the numerator with the degree of the denominator:
- If the degree of the numerator is smaller, the limit at infinity is $0$ (Example 5).
- If the degrees are equal, the limit is the leading coefficient of the numerator divided by the leading coefficient of the denominator (Examples 4 and 7).
- If the degree of the numerator is larger, there is no finite limit. The limit is $\infty$ or $-\infty$ (Example 6).