### Video Transcript

One-Sided Limits

In this video, weβll learn how to
find the value of one-sided limits both graphically and algebraically. As the name implies, one-sided
limits involve approaching a point, letβs say π₯ equals π, from one direction. That is, the positive or the
negative direction. Now at first, it may not seem
apparent why this would be useful to us. But letβs explore further using the
following example.

Consider the function π of π₯ is
equal to π₯ plus one at all points where π₯ is not equal to two.

If we wanted to find the limit as
π₯ approaches two of π of π₯, we could do so by direct substitution. We can do so because even though π₯
equals two is not in the domain of our function, the limit concerns values of π₯
which are arbitrarily close to two but not where π₯ is equal to two. Performing a substitution, we find
our answer is two plus one which is three. Letβs now take a look at our
function πof π₯ on a graph.

We first note that the hollow dot
here tells us that the function π of π₯ is not defined at this point, where π₯ is
equal to two. Letβs now consider the process of
taking a limit but graphically. We know that our limit concerns
values of π₯ which are close to two. So letβs look at a value which is
just slightly smaller; letβs say 1.8. If π₯ is 1.8, the value of our
function is π of 1.8. The value of this is 1.8 plus one
which is of course 2.8.

To get a more accurate limit, we
need π₯ to approach our value of two. Letβs now look at π₯ equals
1.9. In this case, the value of our
function is 2.9. We could continue this process
getting closer and closer to the value of π₯ equals two. Doing so we find that the value of
our function approaches three as we would expect. Here, we note that we have
approached our value of π₯ equals two from the left or from the negative
direction. We could also perform the same
exercise approaching from the right or from the positive direction. In fact, if we were to do so we
would find that our values of π of π₯ would converge to the same thing. The point that weβre illustrating
here is that approaching the value of π₯ equals two from the left or the right seems
to approach the same value of π of π₯.

Letβs now consider what would
happen if we had a different function, letβs say π of π₯, which is defined
piecewise as the following. π of π₯ is π₯ plus one if π₯ is
less than two and π₯ plus two if π₯ is greater than two. Again we know graphically that
these hollow dots tell us that the value of π of π₯ is not defined at these two
points. And in fact the function π of π₯
is not defined where π₯ is equal to two. If we we were to perform the same
exercise as before approaching the value of π₯ equals two from the left, our value
of π of π₯ approaches three as weβve seen. However, now if we approach π₯
equals two from the right or the positive direction, we would see that the values of
π of π₯ seem to be approaching four. This means that as we move towards
π₯ equals two, our values for π of π₯ seem to be approaching two different values
depending on which direction weβre approaching from.

Given this is the case, it does not
make sense for us to assign a value for the limit as π₯ approaches two of π of
π₯. And in fact we say that this limit
does not exist. However, it is still useful for us
to consider what happens as we approach from the left or the right, as this still
provides useful information about our function. When approaching from the different
directions, we have in fact found the left-sided limit and the right-sided limit for
our function π. Now, the difference in notation
here is quite subtle. But here we see that the negative
symbol in the position that a power would usually go tells us that weβre approaching
from the negative direction and the positive symbol tells us that weβre approaching
from the positive direction.

A slightly more formal definition
for our one-sided limits are then as follows. When π of π₯ can be made
arbitrarily close to some value πΏ as π₯ approaches some value π from the left,
that is to say that π₯ is strictly less than π and not equal to π, then we say
that the left-sided limit as π₯ approaches π of π of π₯ is equal to πΏ. When the same conditions are true
but π₯ approaches π from the right, that is to say, π₯ is strictly greater than π
and not equal to π, then we instead say that the right-sided limit as π₯ approaches
π of π of π₯ is equal to πΏ. Here weβve seen that one-sided
limits give us a useful tool for looking at functions using a piecewise function
with a discontinuity as an example. Now letβs look at an algebraic
example where one-sided limits come in handy.

Find the left-sided limit as π₯
approaches π of π of π₯ given that π of π₯ is equal to five π₯ cos five π₯ plus
two sin five π₯ over π₯ if π₯ is greater than zero and less than π over two and
four over two cos nine π₯ plus π if π₯ is greater than π over two and less than
π.

Here, we have a piecewise function
π of π₯ defined over two different intervals. Our function is clearly undefined
when π₯ is less than or equal to zero or greater than or equal to π. We can also note that due to these
strict inequality symbols π of π₯ is also undefined when π₯ is equal to π over
two. Now, our question is asking us for
the left-sided limit as we can see from this minus symbol. This means that weβre approaching a
value of π₯ equals π from the negative direction. And π₯ is strictly less than
π. Even though we know that π₯ equals
π is not in the domain of our function, we can still try and find a limit since the
limits concern values of π₯ which are arbitrarily close to π but not equal to
π.

Now, we know that the values of π₯
that weβre interested in are less than π but very close to this value. Hence, the interval of our function
that weβre interested in is this one where π of π₯ is equal to four over two cos of
nine π₯ plus π. And we can see this from our
inequality. Since this is the interval for
which π₯ is just less than π, we proceed to evaluate our limit as follows. We take a direct substitution of π₯
equals π into our function. Looking at the term cos of nine π,
we recall that cosine is a periodic function and repeats itself over a period of two
π radians. This means that cos of nine π is
equal to cause of π. And of course this is equal to
negative one.

If we perform this substitution, we
find that our answer becomes the following quotient. And we then arrive at an answer of
four over negative two plus π. We have now answered the
question. And we have found the left-sided
limit as π₯ approaches π of the function π of π₯. In some cases, it might be
difficult to sketch your function. And here weβve demonstrated taking
a one-sided limit without plotting our function graphically. Itβs also worth reflecting back on
the fact that since π of π₯ is undefined when π₯ is greater than π and indeed when
π₯ is equal to π, then the right-sided limit as π₯ approaches π of π of π₯ does
not exist. And indeed the normal limit as π₯
approaches π of π of π₯ cannot be said to exist either.

In this case, it only makes sense
for us to assign a value to the left-sided limit as π₯ approaches π of π of π₯
demonstrating how one-sided limits give us increased precision in our mathematical
descriptions. In the example we just saw, values
of π₯ greater than π were not in the domain of our function π. And hence, we said that the
right-sided limit as π₯ approaches π of π of π₯ did not exist. However, we can now look at an
example to illustrate that even if π of π₯ seems to be defined over all values of
π₯, then maybe some cases whether left or the right-sided limit may not exist. And indeed the normal limit in this
case would not exist either. Letβs take a look at one such
example.

Determine the left-sided limit as
π₯ approaches negative nine of π of π₯ and the right-sided limit as π₯ approaches
negative nine of π of π₯ given that π of π₯ is equal to π₯ plus nine if π₯ is less
than or equal to negative nine and one over π₯ plus nine if π₯ is greater than
negative nine.

Here, weβve been given a function
defined piecewise over two intervals. For the left-sided limit, weβre
approaching π₯ equals negative nine from the negative direction. Hence, π₯ is less than negative
nine. For the right-sided limit, weβre
approaching π₯ equals negative nine from the positive direction. And hence, π₯ is greater than
negative nine. Since π₯ equals negative nine is
the point between the two intervals of our piecewise function, for our left-sided
limit will be in the first interval and for our right-sided limit will be in the
second interval. Letβs work on finding the
left-sided limit.

In this case, our function π of π₯
is π₯ plus nine. We can find this limit by direct
substitution of π₯ equals negative nine into our function. Doing so, we find that our answer
is negative nine plus nine which is equal to zero. The left-sided limit as π₯
approaches negative nine of π of π₯ is therefore zero. Now for the right-sided limit, here
our function π of π₯ is one over π₯ plus nine. Again, we tried direct substitution
of π₯ equals negative nine into our function. This time, doing so gives us an
answer of one over zero. And as we know, dividing one by
zero cannot be evaluated to a numerical value. In cases such as this, we say that
the limit does not exist. And so in a strict sense, this is
the answer to our question.

To more fully understand our
result, however, letβs look at a graphical plot of our function. Here, we have sketched our
graph. And we know that, in the interval
where π₯ is less than or equal to negative nine, we have a well behaved
function. And we know that due to the solid
dot here at negative nine π₯ is indeed defined at this point. For the other interval, we know
that as π₯ approaches negative nine, we have a vertical asymptote. This means that the values of π of
π₯ get arbitrarily large. And this is often represented as
infinity. In this sense, it is common to
write that the right-sided limit as π₯ approaches negative nine of π of π₯ is equal
to positive infinity.

A very important distinction to
make here is that we are not saying that infinity takes a numerical value. Nor are we saying that our limit
exists. Rather, weβre expressing that the
limit does not exist in a particular way. We do this because expressing a
limit like this still gives us useful information about our function, as shown by
the graph. Writing the limit in this way even
without the graph would give us a sense that, at negative nine, we have a
discontinuity. And as we approach this value from
the right, the values of π of π₯ get arbitrarily large. To round out the concept of
one-sided limits, itβs useful to understand the distinctions and the relationships
between the value of our function at π₯ equals π, the normal limit of the function
as π₯ approaches π, and, of course, the left and right limits as π₯ approaches
π.

In particular, let us first
formalize the following relationship. If the left-sided limit and the
right-sided limit as π₯ approaches π of a function π of π₯ exist and are equal to
each other, taking some value πΏ, then the normal limit as π₯ approaches π of the
function π of π₯ also exists and is also equal to πΏ. In fact, we could also take this
rule in reverse by saying that if the normal limit as π₯ approaches π exists and is
equal to πΏ, then it will be equal to the value of the left-sided and the
right-sided limits as π₯ approaches π. And these would also be πΏ.

Weβve already touched on this in
previous examples; but here we reiterate. If the left- and the right-sided
limits disagree or do not exist, then it does not make sense for us to say that the
normal limit itself exists. Itβs also worth noting that the
limit of a function as π₯ approaches π may, in fact, be completely independent from
the value of the function itself at the point where π₯ is equal to π. Letβs look at an example of this in
the following question.

Find the following: π of negative
three the left-sided limit as π₯ approaches negative three of π of π₯, the
right-sided limit as π₯ approaches negative three of π of π₯, and the normal limit
as π₯ approaches negative three of π of π₯. This is part a) through d). Next for part e) through h), we
have find π of one, find the left-sided limit as π₯ approaches one of π of π₯, the
right-sided limit as π₯ approaches one of π of π₯, and the normal limit as π₯
approaches one of π of π₯.

Now at first glance, this is a lot
of work. But we see that the first four
parts of our question are very closely related as are the second four parts of our
question. Here, weβre gonna do things in
sections, first, looking at parts a), b), c), and d). Now, weβve been given a graph which
describes the function π of π₯. The first point we note is that the
hollow dots on our graph describe points where π of π₯ does not exist, whereas the
filled dots on our graph describe points where π of π₯ does exist. Looking at the value of π₯ equals
negative three on our graph, we see that there is a hollow dot at the point negative
three zero and a filled dot at the point negative three two. From this, we can say that when π₯
is equal to negative three, π of π₯ is equal to two. In other words, weβve just answered
part a) of our question. And π of negative three is equal
to two.

Next, we move on to the left-sided
limit as π₯ approaches negative three. As we approach we see from our
graph that the value of π of π₯ gets closer and closer to zero. It does not matter here that we
have a hollow dot at negative three zero since the limits concern values of π₯ which
are arbitrarily close to negative three but not equal to three. We can then say that the left-sided
limit as π₯ approaches negative three of π of π₯ is equal to zero. In fact, the same can be said for
the right-sided limit. As π₯ approaches negative three
from the positive direction, π of π₯ also gets closer and closer to zero. This means that the right-sided
limit as π₯ approaches negative three of π of π₯ is also equal to zero.

Now, given that the left- and the
right-sided limits as π₯ approaches negative three both exist and are equal to the
same value, we can use this general rule to say that the normal limit also exists
and is equal to the same value. We, therefore, conclude that the
normal limit as π₯ approaches negative three of π of π₯ is also equal to zero. We have now answered parts a)
through d) of our question. An interesting point to note here
is that even though the value of π of negative three is equal to two, the left, the
right, and the normal limits as π₯ approaches negative three are all equal to
zero. Again, the limits concern values of
π₯ which are close to, but not equal to, negative three.

In this case, the limits are
completely unconcerned with the exact value of π of negative three and only
concerned with the values π₯ takes as it approaches negative three. In fact, we could remove the point
at negative three two entirely from our graph. And doing so would leave all of our
limits completely unchanged even though the value of π of negative three would be
left undefined. Letβs now move on to the next four
parts of our question.

First, we must find the value of π
of one. From our graph we see that there is
a filled dot at the point one, negative two and a hollow dot at the point one,
four. We know that the filled dot is
where π of π₯ is defined. And hence, π of one is equal to
negative two. Next, for the left-sided limit as
π₯ approaches one for π of π₯, we see what happens to our function as we approach
the value where π₯ equals one from the negative direction. Here, itβs clear that the value for
π of π₯ is approaching negative two. And therefore, this is also the
value of our left-sided limit. For the right-sided limit, we see
what happens to the value of π of π₯ as we approach from the positive
direction. And from our graph, the value of π
of π₯ gets closer and closer to four as π₯ approaches one from the positive
direction. This means that our right-sided
limit is equal to four.

Now for the final part of our
question, the normal limit as π₯ approaches one of π of π₯. Here, we know that both the left-
and the right-sided limits exist. However, they take different
values. And therefore, they disagree given
this fact we can conclude that the normal limit does not exist. We have now answered all parts of
our question. This example illustrates a few
different ways in which the value of the function, the left-sided the right-sided,
and the normal limits of a function can relate to each other.

To finish off, letβs go through a
few key points. The left- and right-sided limits of
some function π of π₯ concern the value of π of π₯ as π₯ approaches some value π
from the negative direction and the positive direction, respectively. A more formal definition has been
given below here. If the left- and the right-sided
limits both exist and agree on some value πΏ, then the normal limit will also exist
and be equal to the same value πΏ. We can also sometimes understand
this rule in reverse, making inferences about the left- and the right-sided limits
given the normal limit.

Sometimes, a limit where the normal
or one-sided may be expressed as positive or negative Infinity. In these cases, we are not saying
that infinity, whether positive or negative, takes a numerical value. Nor are we saying the limit
exists. But rather, this is a particular
way of expressing that the limit does not exist because it gives us useful
information about the function. As a final point, we remember that
the one-sided limits can exist when the normal limit does not. And they therefore provide us with
useful additional tools for describing a function, for example, when looking at a
piecewise function with a discontinuity.