On this screen we’re going to examine the limits at positive infinity of some common functions, including sin(x), $e^x$, and $x^n.$ We’ll use the epsilon-strip to help us see whether the limit does or does not exist.

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Reminder:

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 $L – \epsilon \lt f(x) \lt L + \epsilon.$

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$\color{blue}{\text{I. }f(x) = \sin(x)}$

Let’s consider first the function $f(x) = \sin(x),$ graphed below. The horizontal green “possible-limit line” initially has value $?L = -1.3?.$ The question marks are present because we are not saying that value is the actual limit; clearly it is not. You can adjust this possible-limit value by dragging the green dot up or down along the y-axis.

You can also turn on an accompanying epsilon-strip, with $\epsilon = 0.5$ or $\epsilon = 0.1$, by using the buttons beneath the graph. We could of course set epsilon to be smaller still; we’re using only these two values because they’re sufficient for us to reach the conclusion we’re after.

Please turn the “epsilon = 0.5 strip” on now.

Then zoom-out the graph a bit only along the x-axis so you’re seeing larger and larger values of x. (There’s a reminder of how to do that immediately below the graph.) Then ask yourself: as $x \to \infty,$ is there a value for L such that all of the function’s output values fall within the epsilon-strip for sufficiently large x? If so, what is the approximate value of M such that for $x \gt M$ all of the function’s output values lie in the range $L – \epsilon \lt f(x) \lt L + \epsilon$? Try adjusting the possible value of L to help you decide.

       

As you can see, the answer is no: because the sine function continues to oscillate between $-1$ and 1 no matter how large x is, there is no value of L that satisfies the limit condition. Instead, no matter how far out we go, some of the function’s values always fall outside of the given range.

Remember that for the limit to exist and equal L, we must be able to find a value of M for any value of epsilon that we choose. Here, there is no value of L for which this works given our chosen $\epsilon = 0.5$. And this result is sufficient for us to conclude that the limit does not exist (DNE).

Using the “epsilon = 0.1 strip” provides further illustration of the same conclusion.

Conclusion: $\displaystyle{\lim_{x \to \infty} \sin(x) = \text{ DNE} \, \cmark}.$ Instead, the function simply oscillates forever.

You can imagine (or graph for yourself to see) that the same conclusion holds for $f(x) = \cos(x)$:
$\displaystyle{\lim_{x \to \infty} \cos(x) = \text{ DNE} \, \cmark}.$

$\color{blue}{\text{II. }f(x) = e^x}$

Let’s next examine the power function $f(x) = e^x,$ graphed below. This time two epsilon strips are
available to you are $\epsilon = 100$ and $\epsilon = 10.$ Note that these are not small values of $\epsilon$; we’ve chosen them so you can easily see the strip’s width on the graph to reach your conclusion, and then imagine the effect of making epsilon as small as you wish.

Use the strips to decide: is there a value of L you can choose such that as $x \to \infty,$ all of the
function’s output values fall in the range $L – \epsilon \lt f(x) \lt L + \epsilon$?

       

As you can see, since the function’s output values grow without bound as $x \to \infty,$ the answer is no: there is no single value of L that satisfies the limit condition.

Conclusion: $\displaystyle{\lim_{x \to \infty} e^x = \text{ DNE} \, \cmark}.$ If we choose, we can convey additional information about the way in which the limit does not exist by writing $\displaystyle{\lim_{x \to \infty} e^x = \infty \, \cmark}.$ As always, realize that the $\infty$ on the right side of this equation is not a particular value we can point to, and instead simply means that as $x \to \infty,$ the function’s output “increases in the positive direction without bound:” $f(x) \to \infty.$

$\color{blue}{\text{III. }f(x) = x^n,\; n \ge 1}$

Let’s next examine the power function $f(x) = x^3$ for $x \ge 0,$ graphed below. Here the two epsilon strips are $\epsilon = 10$ and $\epsilon = 1.$

Again use the strips to decide: is there a value of L you can choose such that as $x \to \infty,$ all of the function’s output values fall in the range $L – \epsilon \lt f(x) \lt L + \epsilon$?

Since the function’s output values grow without bound as $x \to \infty,$ the answer is no: there is no single value of L that satisfies the limit condition. Hence $\displaystyle{\lim_{x \to \infty} x^3 = \text{ DNE}}.$ If we choose, we can convey additional information about the way in which the limit does not exist by writing $\displaystyle{\lim_{x \to \infty} x^3 = \infty}.$

To explore the set of functions $f(x) = x^n$ for $n \ge 1$ more generally, please open the “Set the function’s exponent” box above and see the effects of changing n. You can probably convince yourself of the general result:
Conclusion: For $n \ge 1,$ $\displaystyle{\lim_{x \to \infty} x^n = \text{ DNE} \, \cmark}.$ Or, to provide more information about the way in which this limit does not exist, $\displaystyle{\lim_{x \to \infty} x^n = \infty \, \cmark}.$

$\color{blue}{\text{IV. }f(x) = x^n,\; 0 \lt n \lt 1}$

Let’s look now at the function $f(x) = x^{0.5} = x^{1/2} = \sqrt{x}.$ This function clearly grows more slowly than those immediately above. What is its limit as $x \to \infty$?

As you can see, even though $f(x) = x^{0.5}$ grows more slowly than any of the functions $x^n$ with $n \ge 1,$ it still grows without bound as $x \to \infty.$ Hence $\displaystyle{\lim_{x \to \infty} x^{0.5} = \text{ DNE} }.$ Or, to provide more information, $\displaystyle{\lim_{x \to \infty} x^{0.5} = \infty }.$

To explore the set of functions $f(x) = x^n$ for $0 \lt n \lt 1$ more generally, please open the “Set the function’s exponent” box above and see the effects of changing n. You can probably convince yourself of the general result for $n \ge 1$:
Conclusion: For $0 \lt n \lt 1$ $\displaystyle{\lim_{x \to \infty} x^n = \text{ DNE} \, \cmark}.$ Or, to provide more information about the way in which this limit does not exist, $\displaystyle{\lim_{x \to \infty} x^n =\infty \, \cmark}.$

Combined Conclusion for Parts III and IV: Combining the conclusion of this and the preceding section, we have for $f(x) = x^n$ where $n \gt 0$, $\displaystyle{\lim_{x \to \infty} x^n = \text{ DNE} \, \cmark}.$ Or, to provide more information about the way in which this limit does not exist, $\displaystyle{\lim_{x \to \infty} x^n = \infty \, \cmark}.$

$\color{blue}{\text{V. }f(x) = C, \text{constant function}}$

Let’s look now at a constant function, $f(x) = 42.$ (We could use any number C there; we just like the number 42.) The values of epsilon available to you are 1, 0.1, and 0.01. Once you reach your initial conclusion, you can probably imagine making epsilon as small as you’d like.

Use the strips to decide: is there a value of L you can choose such that as $x \to \infty,$ all of the
function’s output values fall in the range $L – \epsilon \lt f(x) \lt L + \epsilon,$ no matter how small $\epsilon$ is? If so, what is that single value of L?

As you can see, only the value $L = 42$ satisfies the condition $L – \epsilon \lt f(x) \lt L + \epsilon$ for any $\epsilon \gt 0.$ On the one hand this result is probably completely obvious: what other value could it be? On the other hand, if you understand how this must be the result given our definition of the limit and the limit condition for any epsilon, you are developing an excellent understanding of the concept of “limit” in general. (If so: good work!)

Conclusion: For a constant value C, $\displaystyle{\lim_{x \to \infty}C = C \, \cmark}$

$\color{blue}{\text{VI. }f(x) = x^{-n} = \dfrac{1}{x^n}}, \; n \gt 0$

Finally, let’s look at the function $f(x) = x^{-2} = \dfrac{1}{x^2}.$ Does $\displaystyle{\lim_{x \to \infty}x^{-2}}$ exist? If so, what is its value?

As you can see, only $L = 0$ satisfies the condition $L – \epsilon \lt f(x) \lt L + \epsilon$ for any $\epsilon \gt 0.$ Hence, $\displaystyle{\lim_{x \to \infty}x^{-2} = 0}.$

To explore the set of functions $f(x) = x^{-n} = \dfrac{1}{x^n}$ for $n \gt 0$ more generally, please open the “Set the function’ exponent” box above and see the effects of changing n. You can probably convince yourself of the general result:
Conclusion: For $n \gt 0$ $\displaystyle{\lim_{x \to \infty} x^{-n} = \lim_{x \to \infty} \dfrac{1}{x^n} = 0 \, \cmark}.$

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On the next screen we’ll use the epsilon strip to examine the limit of some functions as $x \to -\infty.$

The Upshot

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Do you have other functions in mind where the epsilon strip helps you decide whether the limit at infinity exists or not? Join the discussion over on the Forum and let us know!

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