We’re now going to explore limits “at infinity,” of course including lots of free practice problems with complete solutions on upcoming screens.

We put the words “at infinity” in quotes because, as we said before:

Infinity is *not* a specific place.

Instead it means “arbitrarily large.” In this context you can think of it as a process: “keep going.”

That is, whatever number you can think of,

you can *always* go to an even larger value to input into the function.

Hence when you see $x \to \infty,$ think: “What happens as *x* grows and Grows and GROWS,” or “grows without bound,” in the positive *x*-direction. You can thus let *x* have whatever large value it needs to in order to meet whatever criteria you set. (We’ll explain what we mean by “criteria” and “whatever large value it needs to be” below.)

Similarly, when you see $x \to -\infty,$ think: “What happens as *x* grows and Grows and GROWS,” or “grows without bound,” in the negative *x*-direction, and can have whatever negative large value it needs to in order to meet whatever criteria you set. (By “negative large value,” we mean the number is negative and its absolute value is large. That is, you’re heading further and further to the left on the *x*-axis.)

Quick sidenote: Buzz Lightyear may strive for “beyond infinity,” but mathematically “to infinity” already means “can be larger still,” and so there is no “beyond.” Sorry, Buzz!

[In case you’re wondering: We actually all love Pixar’s

With this notion of infinity in mind, let’s recast our definition of limits to apply to “as *x* approaches (positive) infinity,” $x \to \infty$:

The limit of the function $f(x)$ as *x* approaches infinity, written

\[\lim_{x \to \infty} f(x) = L\]

is a number *L* (if one exists) such that $f(x)$ is as close to *L* as we want

whenever *x* is sufficiently large and positive.

Similarly, for “*x* approaches negative infinity,” $x \to -\infty$:

The limit of the function $f(x)$ as *x* approaches negative infinity, written

\[\lim_{x \to -\infty} f(x) = L\]

is a number *L* (if one exists) such that $f(x)$ is as close to *L* as we want

whenever *x* is sufficiently large and negative.

Let’s examine one function, $f(x) = \dfrac{1}{x},$ in some detail to develop an understanding of what the above statements mean in practice.

First limit “at positive infinity” illustration: $\displaystyle{ \lim_{x \to \infty} \dfrac{1}{x}}$

As a first example, let’s consider the function $f(x) = \dfrac{1}{x}$ for $x \gt 0.$ Looking at its graph below, what do you think $\displaystyle{\lim_{x \to \infty} \frac{1}{x}}$ equals?

You probably have an informal thought something like, “$f(x)$ gets close to 0 as *x* gets larger and larger, so the limit is 0.” And that is correct:

\[\lim_{x \to \infty} \frac{1}{x} = 0 \quad \cmark \]
As before, we can make that informal reasoning more precise by saying

“We can get as close to $f(x)$ = 0 as we’d like by making *x* sufficiently large.”

For instance, let’s say we tell you that we need the function’s output to be to be within $\pm 0.0001 \left( = \pm \dfrac{1}{10000}\right)$ of 0. That is, we set our “output error,” or “tolerance,” to be $\epsilon = 0.0001.$ Then we ask you to point to the region on the graph where the *x*-values meet this criteria. Where would you point to?

When you have your answer in mind, click the “Zoom for ε = 0.0001” button beneath the graph to reveal the answer visually.

Graph of $f(x) = \dfrac{1}{x}$ versus *x* to examine $\displaystyle{\lim_{x \to \infty}f(x).}$

For what region are the output values within $\pm\epsilon = \pm 0.0001$ of $y=0$?

For what region are the output values within $\pm\epsilon = \pm 0.0001$ of $y=0$?

Perhaps you did the quick calculation in your head: since the function is $f(x) = \dfrac{1}{x},$ to obtain output values smaller than $0.0001 = \dfrac{1}{10000}$ we need to have our input values be $x > 10,000.$ The zoomed-in graph illustrates this visually: The horizontal dashed blue lines indicate the region with *y*-values $0 \pm 0.0001.$ As you can see, when $x \gt 10,000$ the function’s output values are all within $\pm 0.0001$ of 0.

We’ve marked the vertical line $x = 10,000$ with the label “*M* = 10,000″ to indicate that all input-values to the right of this line give us output values in the range we desire. To verify this, you can place your cursor on the curve and see that for $x \gt M = 10,000,$ *all* of the output (*y*) values are indeed within $\pm 0.0001$ of 0. We can thus make the mathematically correct statement for $f(x) = \dfrac{1}{x}:$

We can get within $\epsilon = 0.0001$ of $f(x) = 0$ by taking $x > M = 10,000.$

Traditionally the capital letter *“M”* (or in some books, *“N”*) is used to denote the edge of this region. It’s a capital letter to remind us that we’re looking at LARGE values of *x* for which the statement is true.

Although the result will probably be obvious, let’s press ahead and instead say that we now want the output values of the function to be within $\pm 1 \times 10^{-8}$ of 0. What is our new value of *M*?

When you have your answer in mind, click the “Zoom for ε = 10^(-8)” button underneath the graph below to reveal the answer visually.

Graph of $f(x) = \dfrac{1}{x}$ versus *x* to examine $\displaystyle{\lim_{x \to \infty}f(x).}$

For what region are the output values within $\pm 1 \times 10^{-8}$ of $y=0$?

For what region are the output values within $\pm 1 \times 10^{-8}$ of $y=0$?

As you probably expected, you can see that for $x \gt M = 1 \times 10^{8},$ the function’s output values are indeed within the range $0 \pm 1 \times 10^{-8}.$ And so we can write

We can get within $\epsilon = 1 \times 10^{-8}$ of $f(x) = 0$ by taking $x > M = 1 \times 10^{8}.$

You can imagine that for *any* value of $\epsilon$ we gave you, you could find a value of M that would satisfy the criterion that the output values be within the range $0 \pm \epsilon.$ Indeed, for this particular function, finding the value of *M* is easy: $M = 1/\epsilon.$ Typically, however, the relationship is not so straightforward, and for other functions we won’t state a formula for it (an expression relating $\epsilon$ to *M*) as we just did here for $f(x) = 1/x.$

This discussion illustrates that when we wrote above that the limit $L = \displaystyle{\lim_{x \to \infty} \dfrac{1}{x} = 0 }$, what we were really saying is that for *any* value of $\epsilon$ we give you, you could give us back a value of *M* such for any value of $x > M$ the function’s output values will fall in the range $L \pm \epsilon.$ Said differently, once we’re to the right of $x = M,$ the function will not output *any* values that fall outside of the range $0 \pm \epsilon.$

The reasoning of this example illustrates the key point about limits at infinity:

The limit at infinity exists and equals *L* *if,*

for *any* value of $\epsilon \gt 0$ that we choose,

there is a value of *M* such that for all $x \gt M$

the function’s output values lie in the range $L\, – \epsilon \lt f(x) \lt L + \epsilon.$

In a few screens we’ll look at some functions that have identifiable limits at $x \to \infty,$ as well as some functions whose limits at infinity do not exist. Before then, let’s apply these same ideas to $x \to -\infty.$

First limit “at negative infinity” illustration: $\displaystyle{ \lim_{x \to -\infty} \dfrac{1}{x}}$

Let’s again consider the function $f(x) = \dfrac{1}{x},$ and now consider happens as $x \to -\infty.$ That is, imagine traveling further and further to the left on the graph shown.

You probably have the notion that as *x* increases without bound in the negative direction — that is, taking on larger and larger negative values — the function $f(x) = \dfrac{1}{x}$ approaches 0. And that is correct:

\[\lim_{x \to -\infty} \frac{1}{x} = 0 \quad \cmark \]
As above, what this really means is that we can obtain output values for the function as close to 0 as we desire by choosing values of *x* that are negative and larger in size (absolute value) than some number *M,* where *M* is now also a negative number. For instance, as you can verify using the calculator below:

â€¢ For $\epsilon = 0.0001,$ we have $M = -10,000;$ and

â€¢ for $\epsilon = 10^{-8},$ we have $M = 10^8.$

â€¢ for $\epsilon = 10^{-8},$ we have $M = 10^8.$

Graph of $f(x) = \dfrac{1}{x}$ versus *x* to examine $\displaystyle{\lim_{x \to -\infty}f(x)}$.

For what region are the output values within $\pm \epsilon$ of $y=0$?

For what region are the output values within $\pm \epsilon$ of $y=0$?

We can generalize to the key point about limits at negative infinity:

The limit at negative infinity exists and equals *L* *if,*

for *any* value of $\epsilon \gt 0$ that we choose,

there is a value of *M* such that for all $x \lt M$

the function’s output values lie in the range $L\, – \epsilon \lt f(x) \lt L + \epsilon.$

We’ll continue to explore these ideas by introducing a new tool to examine other functions on the next screen.

- “Infinity” is
*not*a specific place. Instead it means “arbitrarily large,” and you can think of it as the process of entering ever-larger input values into the function. - The limit of the function $f(x)$ as
*x*approaches infinity, written $\displaystyle{}\lim_{x \to \infty} f(x) = L,$

is a number*L*(if one exists) such that $f(x)$ is as close to*L*as we want whenever*x*is sufficiently large and positive. - More precisely, for $x \to \infty$ the limit exists and equals
*L*the function’s output is within $\pm \epsilon$ of*L*for all $x \gt M$ for some number*M.*And for $x \to -\infty,$ the limit exists and equals*L*if the function’s output lies within $L \pm \epsilon$ for all $x \lt M,$ for some (different) number*M.*

We’ve taken our first steps “toward infinity.” What thoughts or questions do you have? Please post on the Forum to join the discussion!