On this screen we’re going to introduce a new tool, “the epsilon strip,” that will help us determine whether a limit at infinity (or negative infinity) exists. To motivate and illustrate the use of the tool, we’re going to use Desmos to consider a single oscillating function. We’ll then use the tool to examine other functions on the next screen.
The single function f that we’re going to consider on this screen is plotted in the graph immediately below, and as you can see, it oscillates forever. Looking at the graph, do you think $\displaystyle{\lim_{x \to \infty} f(x)}$ exists, or not? What feature(s) of the graph lead you to your initial reaction? If you’d like, you can zoom the buttons underneath the graph to quickly zoom out and easily see more of the graph. You can also use the “Center around x = 4000” button to zoom horizontally to an area that illustrates that, even as the function’s output values get closer and closer to 10, they continue to oscillate around that central value.
When you have your answer in mind for yourself about whether the limit exists or not, please continue to the student dialog below.
Two students discuss whether the limit the exists:
Do you have your answers in mind? If not, please pause here to re-read Lakeesha and Misha’s statements and think through for yourself what each is saying. When you’re ready, read on:
We know that Misha’s reasoning matches that of many students who are first learning about limits at infinity — and even if you don’t agree with her, you can probably understand how a smart student could (mis)conclude that the limit doesn’t exist since the function’s output values keep oscillating. It’s a fair initial reaction! But once again, we must go beyond such initial reactions and look to the definition of the limit in order to make a reasoned judgement:
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 $L – \epsilon \lt f(x) \lt L + \epsilon.$
And by this definition, the limit does exist, and $L=10$ as Lakeesha states. We can use the interactive graph below to illustrate this conclusion: We have placed a horizontal green line at the value of the limit, $L = 10.$ And let’s take our initial value of $\epsilon = 1.$ (We’ll make it smaller later.) The purple “$\epsilon$-strip” on the graph shows the region with output values in the region $L – \epsilon \lt f(x) \lt L + \epsilon,$ which is thus currently $9 \lt f(x) \lt 11.$
With the initial zoom-level set, we see many values of $f(x)$ fall outside the desired range. However, if you click the “Zoom Out 1” button beneath the graph, you will see that for values of x greater than approximately $M = 320$, all of the function’s output values fall within the desired range of the purple strip. The “Zoom Out 2” button will zoom out further still.
Use the buttons below to change the value of $\epsilon:$
When you’re ready, you can use the buttons above to change the size of $\epsilon.$ We’ll also automatically zoom the graph to provide a window where it’s easy to see approximately the value of M for the given $\epsilon.$ For instance:
We imagine that you can see the utility of being able to see the epsilon strip in the graph above, and roughly determine, quickly, the value of M for the current $\epsilon$-value.
Another feature of the epsilon strip is that we can use it to find, or check, the value of L itself. For instance, in the interactive graph below we’ve set the value of L to incorrectly be $L \overset{?}{=} 9$. (We can’t include the question mark $\overset{?}{=}$ on the graph, so please place it there in your imagination.) We also set $\epsilon = 0.5.$ As you can see, as x increases, some of (or all, actually) the function’s output values clearly fall outside of the $\epsilon$-strip. That fact shows us immediately that for this function, $L \ne 9:$ there is no value of M such that for all $x \gt M$ the function’s output values fall in the range $8.5 \lt f(x) \lt 9.5.$
You can use the button beneath the graph to set $L \overset{?}{=} 9.8$ instead, and the continue below. . .
And on first glance it looks like this works: as x increases, all of the function’s output values fall within the $\epsilon$-strip. But . . .
Then you remember that the limit must hold for all values of $\epsilon \gt 0,$ no matter how small. So let’s try $\epsilon = 0.1.$ Please tap this button when you’re ready:
Ah: now with $\epsilon = 0.1,$ if you scroll just a bit to the right on the graph you will see that as x increases, some (actually, once again all) of the function’s output values fall outside of the epsilon-strip. Hence the limit cannot be 9.8.
You can imagine that we could continue to try different values of L, and would eventually conclude that the only value that works for any value of $\epsilon \gt 0,$ no matter how small, is $L = 10,$ as illustrated by the interactive graph that proceeded this one.
With this result in mind, let’s restate our definition of the Limit at Infinity, now with an emphasis on L:
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 $L\, – \epsilon \lt f(x) \lt L + \epsilon.$
If the limit exists, it is a single value L that satisfies the above criterion.
On the next screen, we’ll look at some functions for which the limit at infinity exists, and other functions for which it does not.
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