2.6: Limits at Infinity

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Throughout this module, if something does not exist, write DNE in the answer box.

Recap Video

Take a look at the following video which recaps the ideas from the section.

Example Video

Below are two videos showing worked examples.

Example 1:

Example 2:

Problems

The limit $\lim _{x \to \infty } f(x)$ is the $y$ -value the function $y = f(x)$ approaches as $x$ gets larger and larger (if it exists).

A similar definition is made for $\lim _{x \to -\infty } f(x)$ .

If $\lim _{x \to \infty } f(x) =L$ or $\lim _{x \to -\infty } f(x) = L$ , then $y=f(x)$ has a horizontal asymptote at $y=L$ .

Consider the graph of a function $f(x)$ shown below.

Based on the graph, we can say $\lim _{x \to \infty } f(x) = \answer {2}$ .

Based on the graph, we can say $\lim _{x \to -\infty } f(x) = \answer {2}$ .

The graph has

The equation of the horizontal asymptote is $y = \answer {2}$ .

What is the maximum number of horizontal asymptotes a graph can have? $\answer {2}$ .

If $f(x) = \cos (x)$ , then $\lim _{x \to \infty } = \answer {DNE}$ .

The limit does not exist because:

Consider the limit $\lim _{x \to \infty } (x^3-x)$ . The following logic is flawed: $\lim _{x \to \infty } (x^3-x) = \lim _{x \to \infty } x^3 - \lim _{x \to \infty } x = \infty - \infty .$ Why is it that this application of limit laws does not work? Select all that apply.

We can factor $x^3-x$ as $x(x+1)(x-1)$ . As $x \to \infty$ , each term gets really big, so the product is really big. Therefore, we can say $\lim _{x \to \infty } (x^3-x) = \answer {\infty }.$

Consider the function $f(x) = x(x+1)^2(x-1)^2$ . The $y$ -intercept of this function is $(0,\answer {0})$ .

The $x$ -intercepts of this function are $x = 0, -1$ , and $\answer {1}$ .

The limit $\lim _{x \to \infty } f(x) = \answer {\infty }$ .

The limit $\lim _{x \to -\infty } f(x) = \answer {-\infty }$ .

Based on all this information, which of the following looks like it could be the graph of $f(x)$ ?

If $\displaystyle f(x) = \frac {x^3+x+1}{x^3+1}$ , evaluate:

$\lim _{x \to \infty } f(x)$ : Notice as $x$ gets large, both the numerator and denominator of $f(x)$ become larger, so we aren’t quite sure what to make of the ratio. To make it easier for us, we will multiply the top and bottom by $1$ over the highest power of $x$ that occurs in the denominator. In this case, that means $1/x^3$ . If we do this, we get $\frac {x^3+x+1}{x^3+1} \cdot \frac {\frac {1}{x^3}}{\frac {1}{x^3}} = \frac {\answer {1+\frac {1}{x^2}+\frac {1}{x^3}}}{1+\frac {1}{x^3}}.$ Now, notice that as $x \to \infty$ , the numerator approaches $\answer {1}$ , as does the denominator. Therefore, $\lim _{x \to \infty } f(x) = \frac {1}{1} = \answer {1}.$
$\lim _{x \to -\infty } f(x)$ : In this case, all the algebra from the previous part remains the same, so in this case, $\lim _{x \to -\infty } f(x) = \lim _{x \to -\infty } \frac {1+\frac {1}{x^2}+\frac {1}{x^3}}{1+\frac {1}{x^3}} = \answer {1}.$

Consider $f(x) = \frac {x^2+\sqrt {x}+1}{2x^2+3}$ . Using similar steps as in the previous problem, we can say $\lim _{x \to \infty } f(x) = \answer {1/2}$ .

We can also compute $\lim _{x \to -\infty } f(x) = \answer {1/2}$ .

How many horizontal asymptotes does the graph of $y = f(x)$ have? $\answer {1}$ .

Consider $f(x) = \frac {-x^3+1}{x^2+x}$ . Then $\lim _{x \to \infty } f(x) = \answer {-\infty }$ .

We can also compute $\lim _{x \to -\infty } f(x) = \answer {\infty }$ .

How many horizontal asymptotes does the graph of $y = f(x)$ have? $\answer {0}$ .

Evaluate $\lim _{x \to \infty } \frac {\sqrt {9x^2+1}}{2x+3}$ .

We will repeat the steps of the previous two problems. Namely, we will multiply by $1$ over the highest power of $x$ that appears in the denominator, meaning $1/x$ in this case. Doing this gives us $\frac {\sqrt {9x^2+1}}{2x+3} = \frac {\sqrt {9x^2+1} \cdot \frac {1}{x}}{(2x+3) \cdot \frac {1}{x}}.$ In the denominator, we can just multiply through by $1/x$ . In the numerator, the trick is to write $x = \sqrt {x^2}$ (which is true because $x > 0$ as $x \to \infty$ ). Therefore, $\frac {1}{x} = \frac {1}{\sqrt {x^2}} = \sqrt {\frac {1}{x^2}}.$ When we write it this way, namely $\frac {\sqrt {9x^2+1} \cdot \sqrt {\frac {1}{x^2}}}{(2x+3) \cdot \frac {1}{x}},$ we can now multiply everything through to get $\frac {\sqrt {\answer {9+\frac {1}{x^2}}}}{2+\frac {3}{x}}.$ Taking a limit as $x \to \infty$ , we get an answer of $\lim _{x \to \infty } f(x) = \frac {\sqrt {9}}{2} = \answer {3/2}.$

Using the same trick, we can get $\lim _{x \to -\infty } \frac {\sqrt {9x^2+1}}{2x+3} = \answer {-3/2}.$ .

Since $x<0$ as $x \to -\infty$ , we have to write $x = -\sqrt {x^2}$ .

The answer should be $-3/2$ .

The $\lim _{x \to \infty } (\sqrt {x^4+5}-x^2) = \answer {0}$ .

Multiply the function by $\frac {\sqrt {x^4+5}+x^2}{\sqrt {x^4+5}-x^2}$ .

The limit $\lim _{x \to -\infty } (\sqrt {x^2+5}-x) = \answer {\infty }$ .

There is no algebra to be done here, since the square root term gets very large and so does $-x$ . When you add these two things together, you still get a very large number.

Evaluate the limit $\lim _{x \to \infty } \frac {\sin ^3(x)}{x^2}$ .

Notice that as $x \to \infty$ , the numerator doesn’t even have a limit. We will get around this by using the squeeze theorem. Again, our problem is the $\sin ^3(x)$ term, so we will first bound that. Since $-1 \leq \sin (x) \leq 1,$ we can say $-1 \leq \sin ^3(x) \leq 1.$ Now we want to make the middle term look like our function, so we will divide everything by $\answer {x^2}$ . Doing this gives $\answer {-\frac {1}{x^2}} \leq \frac {\sin ^3(x)}{x^2} \leq \frac {1}{x^2}.$ Taking a limit as $x \to \infty$ , we see $\lim _{x \to \infty } -\frac {1}{x^2} = \lim _{x \to \infty } \frac {1}{x^2} = \answer {0}.$ Therefore, our limit is $\answer {0}$ .

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