4.4 Limits at Infinity

In the previous section, we shortly touched on the limit as $x$ approaches positive and negative infinity to examine the behavior of a function. Let $f(x)$ be a function with a domain of $(-\mathrm{\infty },\mathrm{\infty })$. In general, we use the notation

$\underset{x\to \mathrm{\infty }}{lim}f(x)=L$

This means that the output of $f(x)$ gets arbitrarily close to $L$ as $x$ becomes larger and larger. We say that the limit of $f(x)$ is $L$ as $x$ approaches infinity. Similarly,

$\underset{x\to -\mathrm{\infty }}{lim}f(x)=L$

This means that the output of $f(x)$ gets arbitrarily close to $L$ as $x$ becomes smaller and smaller. We say that the limit of $f(x)$ is $L$ as $x$ approaches negative infinity.

Consider the following function:

$f(x)=x^2+2x-10$

What is the limit of the function as $x$ approaches $\mathrm{\infty }$ and $-\mathrm{\infty }$? As $x$ becomes larger and larger, the outcomes of the function $f(x)$ get larger and larger. On the other hand, as $x$ becomes smaller and smaller, the outcomes of the function $f(x)$ get larger and larger. Therefore, we have:

$\underset{x\to \mathrm{\infty }}{lim}{x}^2+2x-10=\mathrm{\infty }$

$\underset{x\to -\mathrm{\infty }}{lim}{x}^2+2x-10=\mathrm{\infty }$

How about the following function?

$f(x)=\frac{1}{x^2+2x-10}$

From the previous example, we know that the denominator has the limit of $\mathrm{\infty }$ as $x$ approaches $\mathrm{\infty }$ and $-\mathrm{\infty }$. This means that the output of the function gets smaller and goes to 0 as $x$ approaches $\mathrm{\infty }$ and $-\mathrm{\infty }$.

$\underset{x\to \mathrm{\infty }}{lim}\frac{1}{x^2+2x-10}=0$

$\underset{x\to -\mathrm{\infty }}{lim}\frac{1}{x^2+2x-10}=0$

One important thing to remember is that if $n$ is a rational number, then

$\underset{x\to \mathrm{\infty }}{lim}\frac{1}{x^n}=0$

Moreover, if $n$ is a rational number such that $x^n$ is defined for all $x$, then

$\underset{x\to -\mathrm{\infty }}{lim}\frac{1}{x^n}=0$

Now, let’s think about finding the limit of rational functions. Suppose we have the following function:

$f(x)=\frac{x^2+3x-1}{x^3-3x^2+10x-10,000}$

What is the limit of the function as $x$ approaches infinity? You might apply one of the properties from the previous section to find the limit of a quotient.

$\begin{array}{rl}\underset{x\to \mathrm{\infty }}{lim}f(x)& =\underset{x\to \mathrm{\infty }}{lim}\frac{x^2+3x-1}{x^3-3x^2+10x-\mathrm{10,000}}\\ & =\frac{\underset{x\to \mathrm{\infty }}{lim}{x}^2+3x-1}{\underset{x\to \mathrm{\infty }}{lim}{x}^3-3x^2+10x-\mathrm{10,000}}\\ & =\frac{\mathrm{\infty }}{\mathrm{\infty }}\end{array}$

So should we say that the limit is infinity? Or since it is infinity divided by infinity, is the limit 0? When we have a rational function, we divide both the numerator and the denominator of the function by the highest power of $x$ in the denominator of the fraction. In this function, the highest power is $x^3$. So, we get

$\begin{array}{rl}\underset{x\to \mathrm{\infty }}{lim}f(x)& =\underset{x\to \mathrm{\infty }}{lim}\frac{x^2+3x-1}{x^3-3x^2+10x-\mathrm{10,000}}\\ & =\underset{x\to \mathrm{\infty }}{lim}\frac{\frac{x^2+3x-1}{x^3}}{\frac{x^3-3x^2+10x-\mathrm{10,000}}{x^3}}\\ & =\underset{x\to \mathrm{\infty }}{lim}\frac{\frac{1}{x}+\frac{3}{x^2}-\frac{1}{x^3}}{1-\frac{3}{x}+\frac{10}{x^2}-\frac{\mathrm{10,000}}{x^3}}\\ & =\frac{\underset{x\to \mathrm{\infty }}{lim}(\frac{1}{x}+\frac{3}{x^2}-\frac{1}{x^3})}{\underset{x\to \mathrm{\infty }}{lim}(1-\frac{3}{x}+\frac{10}{x^2}-\frac{\mathrm{10,000}}{x^3})}\\ & =\frac{\underset{x\to \mathrm{\infty }}{lim}\frac{1}{x}+\underset{x\to \mathrm{\infty }}{lim}\frac{3}{x^2}-\underset{x\to \mathrm{\infty }}{lim}\frac{1}{x^3}}{\underset{x\to \mathrm{\infty }}{lim}1-\underset{x\to \mathrm{\infty }}{lim}\frac{3}{x}+\underset{x\to \mathrm{\infty }}{lim}\frac{10}{x^2}-\underset{x\to \mathrm{\infty }}{lim}\frac{\mathrm{10,000}}{x^3}}\\ & =\frac{0+0-0}{1-0+0-0}\\ & =0\end{array}$

You can try finding the limit of the same function as $x$ approaches negative infinity. You should get 0 as the limit.

Let’s practice finding limits with the following expressions:

$\underset{x\to \mathrm{\infty }}{lim}\frac{10x+4}{-2x-10}$

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

$\underset{x\to \mathrm{\infty }}{lim}\frac{x+3}{\sqrt{4x^2+1}}$

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

$\underset{x\to \mathrm{\infty }}{lim}\frac{e^x+1}{e^{2x}+10}$

Before you scroll down, you may want to take time to practice and then compare your answer with the following solution. We will solve from top to the bottom, one by one.

Here is the first one.

$\begin{array}{rl}\underset{x\to \mathrm{\infty }}{lim}\frac{10x+4}{-2x-10}& =\underset{x\to \mathrm{\infty }}{lim}\frac{\frac{10x+4}{x}}{\frac{-2x-10}{x}}\\ & =\underset{x\to \mathrm{\infty }}{lim}\frac{10+\frac{4}{x}}{-2-\frac{10}{x}}\\ & =\frac{\underset{x\to \mathrm{\infty }}{lim}(10+\frac{4}{x})}{\underset{x\to \mathrm{\infty }}{lim}(-2-\frac{10}{x})}\\ & =\frac{\underset{x\to \mathrm{\infty }}{lim}10+\underset{x\to \mathrm{\infty }}{lim}\frac{4}{x}}{\underset{x\to \mathrm{\infty }}{lim}-2-\underset{x\to \mathrm{\infty }}{lim}\frac{10}{x}}\\ & =\frac{10+0}{-2-0}\\ & =\frac{10}{-2}\\ & =-5\end{array}$

Here is the second.

$\begin{array}{rl}\underset{x\to -\mathrm{\infty }}{lim}\frac{3-2x^2}{4x^3+2x}& =\underset{x\to -\mathrm{\infty }}{lim}\frac{\frac{3-2x^2}{x^3}}{\frac{4x^3+2x}{x^3}}\\ & =\underset{x\to -\mathrm{\infty }}{lim}\frac{\frac{3}{x^3}-\frac{2x^2}{x^3}}{\frac{4x^3}{x^3}+\frac{2x}{x^3}}\\ & =\underset{x\to -\mathrm{\infty }}{lim}\frac{\frac{3}{x^3}-\frac{2}{x}}{4+\frac{2}{x^2}}\\ & =\frac{\underset{x\to -\mathrm{\infty }}{lim}(\frac{3}{x^3}-\frac{2}{x})}{\underset{x\to -\mathrm{\infty }}{lim}(4+\frac{2}{x^2})}\\ & =\frac{\underset{x\to -\mathrm{\infty }}{lim}\frac{3}{x^3}-\underset{x\to -\mathrm{\infty }}{lim}\frac{2}{x}}{\underset{x\to -\mathrm{\infty }}{lim}4+\underset{x\to -\mathrm{\infty }}{lim}\frac{2}{x^2}}\\ & =\frac{0-0}{4+0}\\ & =0\end{array}$

Here is the third.

$\begin{array}{rl}\underset{x\to \mathrm{\infty }}{lim}\frac{x+3}{\sqrt{4x^2+1}}& =\underset{x\to \mathrm{\infty }}{lim}\frac{\frac{x+3}{\sqrt{x^2}}}{\frac{\sqrt{4x^2+1}}{\sqrt{x^2}}}\\ & =\underset{x\to \mathrm{\infty }}{lim}\frac{\frac{x}{x}+\frac{3}{x}}{\sqrt{\frac{4x^2}{x^2}+\frac{1}{x^2}}}\\ & =\underset{x\to \mathrm{\infty }}{lim}\frac{1+\frac{3}{x}}{\sqrt{4+\frac{1}{x^2}}}\\ & =\frac{\underset{x\to \mathrm{\infty }}{lim}(1+\frac{3}{x})}{\underset{x\to \mathrm{\infty }}{lim}\sqrt{4+\frac{1}{x^2}}}\\ & =\frac{\underset{x\to \mathrm{\infty }}{lim}1+\underset{x\to \mathrm{\infty }}{lim}\frac{3}{x}}{\sqrt{\underset{x\to \mathrm{\infty }}{lim}4+\underset{x\to \mathrm{\infty }}{lim}\frac{1}{x^2}}}\\ & =\frac{1}{\sqrt{4}}\\ & =\frac{1}{2}\end{array}$

For this one, we want to remember that

$\sqrt{x^2}=(x^2{)}^{\frac{1}{2}}=x$

Let’s move on to the fourth.

$\begin{array}{rl}\underset{x\to \mathrm{\infty }}{lim}\frac{x^4+5x^2}{3x^3-2x^2+x-1}& =\underset{x\to \mathrm{\infty }}{lim}\frac{\frac{x^4+5x^2}{x^3}}{\frac{3x^3-2x^2+x-1}{x^3}}\\ & =\underset{x\to \mathrm{\infty }}{lim}\frac{x+\frac{5}{x}}{3-\frac{2}{x}+\frac{1}{x^2}-\frac{1}{x^3}}\\ & =\frac{\underset{x\to \mathrm{\infty }}{lim}(x+\frac{5}{x})}{\underset{x\to \mathrm{\infty }}{lim}(3-\frac{2}{x}+\frac{1}{x^2}-\frac{1}{x^3})}\\ & =\frac{\underset{x\to \mathrm{\infty }}{lim}x+\underset{x\to \mathrm{\infty }}{lim}\frac{5}{x}}{\underset{x\to \mathrm{\infty }}{lim}3-\underset{x\to \mathrm{\infty }}{lim}\frac{2}{x}+\underset{x\to \mathrm{\infty }}{lim}\frac{1}{x^2}-\underset{x\to \mathrm{\infty }}{lim}\frac{1}{x^3}}\\ & =\frac{\mathrm{\infty }+0}{3-0+0-0}\\ & =\mathrm{\infty }\end{array}$

Here is the last one. We want to remember that

$a^{nx}=a^n\cdot a^x$

Let’s tackle the problem now.

$\begin{array}{rl}\underset{x\to \mathrm{\infty }}{lim}\frac{e^x+1}{e^{2x}+10}& =\underset{x\to \mathrm{\infty }}{lim}\frac{e^x+1}{e^2\cdot e^x+10}\\ & =\underset{x\to \mathrm{\infty }}{lim}\frac{\frac{e^x+1}{e^x}}{\frac{e^2\cdot e^x+10}{e^x}}\\ & =\underset{x\to \mathrm{\infty }}{lim}\frac{1+\frac{1}{e^x}}{e^2+\frac{10}{e^x}}\\ & =\frac{\underset{x\to \mathrm{\infty }}{lim}(1+\frac{1}{e^x})}{\underset{x\to \mathrm{\infty }}{lim}(e^2+\frac{10}{e^x})}\\ & =\frac{\underset{x\to \mathrm{\infty }}{lim}1+\underset{x\to \mathrm{\infty }}{lim}\frac{1}{e^x}}{\underset{x\to \mathrm{\infty }}{lim}{e}^2+\underset{x\to \mathrm{\infty }}{lim}\frac{10}{e^x}}\\ & =\frac{1+0}{e^2+0}\\ & =\frac{1}{e^2}\end{array}$