On this screen, we’re going to look quickly at some functions where the limit exists at a particular point, and then consider some where the limit does not exist.

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Reminder: \[\text{The limit of the function $f(x)$ as $x$ approaches $a$, written} \] \[\lim_{x \to a} f(x) = L \] \[\text{is a number $L$ (if one exists)}\] \[\text{such that $f(x)$ is $\underbrace{\text{as close to}}_{\Large{\text{within } \pm
\epsilon} \text{ of}}$ $L$ as we want whenever $x$ is $\underbrace{\text{sufficiently close to}}_{\Large{\text{within }
\pm \delta \text{ of}}}$ $a.$ }\]

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A few limits that do exist

We’re going to look at several functions where the limit does exist, with an emphasis on how we know the limit exists. While the result that “the limit exists” for each of these functions may seem obvious and trivial, we need to lay this groundwork in order to recognize when a limit does not exist, both further down on this screen, and in your homework and exam questions.

Function 1: $f(x) = 6 – (x-5)^2$ as $x \to 4$

Let’s first consider $\displaystyle{\lim_{x \to 4}\left[ 6 – (x-5)^2 \right] = 5}.$ (There’s nothing special about the point $x=4;$ we could have chosen any point of interest.)

The graph of this function is the upside-down parabola shown. We’re focused on the limit as $x \to 4$: as you can see, you can get as close to the limit value $L=5$ as you’d like by being sufficiently close to $x=4.$ In fact in this case the function is defined at $x=4,$ with $f(4) = L,$ but that’s irrelevant: focusing on the limit, we care only that we can get as close to $f(x)=L$ as we’d like by being sufficiently close to $x=4.$

Let’s examine another function at a different point of interest.

Function 2: $g(x) = \dfrac{\sin(x)}{x}$ as $x \to 0$

Consider $\displaystyle{\lim_{x \to 0}\dfrac{\sin(x)}{x} = 1.}$

The function $g(x) = \dfrac{\sin(x)}{x}$ is not defined for $x=0$, but you can see visually that $L=1$ … meaning, of course, that you can get as close to $L=1$ as you’d like by being sufficiently close to $x=0,$ as long as you’re not actually at $x=0.$

Let’s consider another function at yet another point of interest.

Function 3: $h(x) = \dfrac{\sqrt{x+1}-2}{x-3}$ as $x \to 3$

Consider now $\displaystyle{\lim_{x \to 3}\dfrac{\sqrt{x+1}-2}{x-3} = L.}$

This function is not defined for $x=3,$ but we can see that the limit exists, even if we don’t know immediately what its value is (although we can see visually it is around 0.25). The fact that we don’t know the value is of no consequence currently: we care only that the limit exists, since we can get as close to the value L as we’d like by being sufficiently close to $x=3.$ (We could, if we cared to, certainly approximate its value to within whatever error bounds we choose, by using the technique of the lab on the preceding screen.)

Some Limits that Exist, Summary
The key thing to notice in each cases above is that you can literally point to the place on the screen where the limiting value is, even if you don’t know exactly what its value is.

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A few limits that do not exist

Let’s now introduce some cases where the limit is does not exist, a phrase that is frequently abbreviated as DNE (Does Not Exist). On the coming screens we’ll dive more deeply into why these limits don’t exist; for now, let’s take a quick overview:

Function 4: Disjointed piecewise function illustrates a “jump”

Consider this function that “jumps” in value at $x=0:$
\[t(x) =
\begin{cases}
-3 & \text{for } x \lt 0 \\
\phantom{-}5 & \text{for } x \ge 0
\end{cases}\]

This function “jumps” at $x=0$ from $-3$ to 5, and hence the limit at $x=0$ does not exist: there is no single number L that we can be as close to as we’d like for values of x both less than and greater than $x=0.$ That is, there is no single place on the graph you can point to that represents the limit at $x=0.$

Notice that at all other points the limit does exist. For example, considering just two particular points: $\displaystyle{\lim_{x \to -4}t(x) = -3},$ and $\displaystyle{\lim_{x \to 2}t(x) = 5}.$

But at $x=0$:
\[\lim_{x \to 0}t(x) = \text{DNE} \]

Function 5: $j(x) = \sin(1/x)$ as $x \to 0$ illustrates infinite oscillations

Let’s first think this through as if we didn’t have the graph to rely on, since on an exam you might not be provided with one or the means to create one.

This function is not defined for $x=0,$ which in and of itself tells us nothing about the limit there. Instead, to think about the limit we must think about what happens as x takes on values closer and closer to 0. We can reason as follows:

First, the inner function, $\dfrac{1}{x},$ becomes larger and larger in size, growing and growing and growing (1) in the positive direction if we’re on the right side of 0, so for $x \gt 0,$ and (2) in the negative direction if we’re on the left side of 0, for $x \lt 0.$ The key point is that $1/x$ never approaches a single value as $x \to 0$; instead, its value keeps changing, growing and growing in size (in either the positive or negative direction).

Second, recall that the sine function oscillates between $-1$ and 1:
\[-1 \le \sin(\text{whatever}) \le 1 \quad \text{Always.}\] That means that as $1/x$ grows and grows (either positively or negatively) as $x \to 0$, always changing values, the sine function keeps oscillating between $-1$ and 1. As it turns out, $\sin (1/x)$ oscillates faster and faster the closer we get to $x=0,$ as you can see in the graph above. No matter how close you get to $x=0,$ there are some values of x with function output values equal to 1, other values of x with function output values equal to $-1,$ and all other values lie between those two extremes.

Hence there is no single value L for the function’s output value that we can get as close to as we’d like by being sufficiently close to $x=0,$ and the limit does not exist (DNE):

\[ \lim_{x \to 0} \sin(1/x) = \text{ DNE} \]

Of course we can reach this conclusion much more quickly by relying on the Desmos graph: if you move the blue dot near $x=0$ from both the left and the right sides, you will see that there is no single value L we can be as close to as we’d like by being sufficiently close to $x=0.$ Instead, as we move closer and closer, the values continue to oscillate around — and hence the limit DNE.

Function 6: $r(x) = \dfrac{1}{(x-2)^4}$ as $x \to 2$ illustrates “infinite limit” (DNE)

The function $r(x) = \dfrac{1}{(x-2)^4}$ is not defined for $x=2,$ which – once again – in and of itself tells us nothing about the limit there. But, we see that as we get closer and closer to $x=2,$ the values of the function become larger, and larger, and larger still . . . without bound. There is no single place you can point to on the graph that we can get as close to as you’d like, and hence the limit does not exist.

You may be thinking that “the limit as $x=2$ is infinity,” and if so you are correct. But we must be very clear: infinity is not a number or a particular place. Instead, when we say “infinity” in this context, what we mean is “arbitrarily large.” Whatever large positive number you can think of, this function will output that value if you’re close enough to $x=2.$ And if you want an even bigger y-value, then simply move closer still to $x=2$ and the function will output it. That’s what we mean by “the limit is infinity”: the function’s output can be made as large as we’d like, without bound.

So both of these statements are true:

When you see the limit for a function like this, you’re like to encounter both $\displaystyle{ \lim_{x \to 2}r(x)} = \text{DNE}$ and $\displaystyle{ \lim_{x \to 2}r(x)} = \infty$ because both are correct. The second simply conveys more information than the first, specifying the way in which the limit does not exist.

Some Limits that Do Not Exist: Summary
We’ve seen three ways in which a limit does not exist at $x=a$:

Questions

Let’s consider a few functions for which students often have conflicting thoughts regarding whether the limit exists or not.

Question 1 illustrates an important point: We mentioned earlier that mathematicians took hundreds of years to settle on the definition of “limit” that we now use. Part of the challenge of developing a usable definition is that it must be applicable to any function, and return a result that is useful in our larger framework. Indeed, it would be problematic (as we’ll see later) if Michael were correct and this limit did not exist: we would have a whole separate way of dealing with constant functions instead of the framework we’ll use for every other function. Instead, the definition was developed to return a useful value (the only possible value) for constant functions, just as it does for many other situations we will examine.

Let’s consider another problem.

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We’ll examine two of the ways limits don’t exist on the upcoming screens, starting on the next screen with the function above that “jumps” from one value to another.

For now, what questions or comments do you have about the material on this screen? Please let your fellow community members, including us, know over on the Forum!

The Upshot

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