### Video Transcript

Classifying Discontinuities

In this lesson, weβll learn how to
identify the different types of function discontinuity at a given point. When thinking about discontinuities,
itβs first helpful to recap the condition for continuity at a point. This is that the limit as π₯
approaches π of some function π of π₯ is equal to the same function, evaluated
when π₯ equals π, π of π. Of course, the implications of this
are that the left and the right limits as π₯ approaches π of π of π₯ must exist
and agree and that π must be in the domain of the function. So π of π must be defined. When the function π of π₯ does not
satisfy the continuity condition, we say that our function is not continuous. If this happens at a point, we say
that we have a discontinuity. Let us now examine the various
different ways in which this can happen.

Our first case is that of a
removable discontinuity. This occurs when the limit as π₯
approaches π of π of π₯ exists and is finite. However, π of π is not equal to
the value of this limit. To illustrate this type of
discontinuity, here is the example function π one of π₯. We can clearly see that π₯ equals
three is not in the domain of our function. And this is shown by the hollow dot
on our graph at this point. π one of three is not defined. And hence this is our removal
discontinuity. Before proceeding, let us look at
another example, function π two, which is defined in the same way as π one when π₯
is not equal to three but is defined as one when π₯ is equal to three. Now for both π one and π two, the
limit as π₯ approaches three is equal to four.

In the case of π one, our function
was undefined when π₯ equals three. In the case of π two, our function
is equal to one when π₯ equals three. This is denoted by the solid dot on
our graph. In both cases, however, this is not
equal to the value of the limit. This illustrates that our function
may or may not be defined at the point where our removable discontinuity exists. A final tip which might help you
remember, these discontinuities are named as such because we could remove the
discontinuity by redefining the function at a single point, in both of these cases,
the point where π₯ is equal to three.

The next type of discontinuity weβll
look at is called an essential discontinuity. And this is sometimes called a
nonremovable discontinuity. These occur when either the left,
the right, or both of the one-sided limits as π₯ approaches π of π of π₯ do not
exist. Here, it is worth remembering that
when we say that a limit is equal to a positive or a negative infinity, this is
simply a particular way of expressing that the limit does not exist. Expressing the limit in this way
does, however, give us useful information about our function. So weβll add it as a side note to
our definition.

To illustrate, letβs look at an
example function π three, which is one over π₯. As π₯ approaches zero from the
left, our function will approach a negative infinity. And as π₯ approaches zero from the
right, our function will approach positive infinity. Again, since infinity is a concept
and not a number, these limits do not exist. This satisfies our criteria and we
therefore have an essential discontinuity at π₯ equals zero. Itβs worth noting that even if we
used a different example function, π four, in which the left- and the right-sided
limits approach the same infinity, which in this case is positive infinity, and we
were able to say that the normal limit as π₯ approaches zero is equal to
infinity. This still satisfies that criteria
for an essential discontinuity at π₯ equals zero, since none of these limits
exist.

Let us now look at a different
example function. In the case of a function such as
sin of one over π₯ as π₯ approaches zero from either the left or the right, the
value of the function itself will oscillate more and more rapidly between negative
one and one. Given this information, it doesnβt
make sense for us to assign a value to either the left- or the right-sided limits,
since π five of π₯ seems to be approaching two values simultaneously. Since we say that these limits do
not exist, this again satisfies that criteria for an essential discontinuity at π₯
equals zero. As a final note, although both of
these cases are indeed essential discontinuities, we sometimes use more specific
language when referring to them, calling the first case an infinite discontinuity
and the second case an oscillating discontinuity.

The final case weβll look at is
called a jump discontinuity. This occurs when both of the
one-sided limits as π₯ approaches π of π of π₯ exist and are finite but are not
equal to each other. You will often, but not always, see
a jump discontinuity when a function defined piecewise, as with the example function
π six shown here. If we examine the boundary between
our two subfunctions, which occurs when π₯ equals two, we can clearly see that as we
approach from the left, π of π₯ approaches three and as we approach from the right,
π of π₯ approaches two.

We might also note the position of
the hollow dot and the filled dot which tells us that π of two is defined here and
is equal to two. But actually for a jump
discontinuity, weβre not particularly interested in this fact. Since both of our one-sided limits
exist and are finite but are not equal to each other, weβve already satisfied the
condition for a jump discontinuity. In fact, this would be the same
case, even if π of two were undefined, the more important feature being the jump
that we observe at π₯ equals two. Okay, now that weβve seen the
different types of discontinuity, let us look at an example.

Find the type of discontinuity that
the function π has at π₯ equals zero, π₯ equals two, π₯ equals five, and π₯ equal
six if it has any discontinuity at these points.

For this question, weβve been given
a graph and asked to identify whether the discontinuities exist at certain points
and to classify them if so. To answer this question, let us go
through the different types of discontinuity that weβre aware of, specifically the
features of each that we would observe on a graph. The first type of discontinuity
that we know is a removable one. This is when the limit as π₯
approaches π of π of π₯ exists and is finite, equaling some value πΏ here. But π of π is not equal to this
value, πΏ. Looking at our graph, we see that
as π₯ approaches two from the left and from the right, π of π₯ approaches one. In other words, the limit as π₯
approaches two of π of π₯ is equal to one.

Another thing we notice when π₯
equals two is that π of two is not defined to be one, denoted by the hollow dot,
but rather is defined to be negative one, denoted by the solid dot. In other words, π of two is equal
to negative one. Weβve now found that the limit as
π₯ approaches two of π of π₯ is not equal to π of two. And this is the condition for a
removable discontinuity. We have, therefore, answered part two
of the question. Let us now move on to the next type
of discontinuity, an essential this continuity. This occurs when either the left,
the right, or both of the one-sided limits as π₯ approaches π of π of π₯ do not
exist. Looking at the graph, we see that
as π₯ approaches six from both the left and the right, we see that the value of π
appears to approach negative infinity.

Here we recall that when we say
that a limit is equal to infinity, whether positive or negative, this is just a
particular way of expressing that the limit does not exist, since infinity is a
concept instead of a number. Given this information, we have
satisfied the condition for an essential discontinuity, which is that at least one
of our one-sided limits must not exist. And in this case, in fact, both do
not. We have therefore found that, at π₯
equal six, we have an essential discontinuity. Finally, let us think about jump
discontinuities. These occur when both of the
one-sided limits as π₯ approaches π of π of π₯ exist and are finite but are not
equal to each other. Again, we look at our graph to find
the cases where this might be true.

Observing π₯ equal zero, we see that
as π₯ approaches from the left, the value of π also approaches zero, whereas as π₯
approaches zero from the right, the value of π approaches three. Both of our one-sided limits as π₯
approaches zero of π of π₯ do exist and are finite. However, they are not equal to each
other, which is the condition for a jump discontinuity. We, therefore, found that π has a
jump discontinuity when π₯ equals zero. To finish off this question, we
must evaluate the point where π₯ is equal to five. As π₯ approaches five from both the
left and the right, the value of π approaches three. Since both of the one-sided limits
both exist and agree, we can also say the normal limit exists and takes this same
value.

Another thing we can see is that π
of five is defined to be three, as seen by the solid dot on our graph. Now, with these two bits of
information, we have found that the limit as π₯ approaches five of π of π₯ is equal
to π of five. You may recognize this as the
continuity condition. And weβve, therefore, proved that
π is continuous when π₯ is equal to five. We do have a sharp corner at this
point, which means that π is not differentiable. However, this is outside the scope
of this video. Since weβve proved continuity here,
by definition, we can say that π has no discontinuity when π₯ equals five. With this information, weβve
completed the question and weβve identified all the discontinuities shown on the
graph of π.

Okay, weβve seen some graphical
examples of discontinuities. But now letβs consider algebraic
examples where no visual cues are given. The first thing we should be aware
of is when dealing with a piecewise function, it is always worthwhile to check and
evaluate the boundaries between the different subfunctions. Letβs have a look at an example of
this now.

Consider the function π of π₯ is
equal to one minus π₯ when π₯ is less than zero, zero when π₯ equals zero, and one
plus two π₯ when π₯ is greater than zero. Part one, what is π of zero?

For this question, weβve been given
a piece wise function defined by three different subfunctions. To begin, we simply must evaluate
π when π₯ equals zero. In fact, the second branch of our
subfunction defines this telling us that π is zero when π₯ equals zero. We can, therefore, simply state
that πof zero is equal to zero. And we have answered the first part
of our question.

Part two, what is the limit as π₯
approaches zero from the left of π of π₯?

Now for this part of the question,
since π₯ is approaching zero from the left, we know that π₯ is less than zero. And hence, π of π₯ is defined by
our first subfunction, one minus π₯. When finding our limit, we can,
therefore, replace π of π₯ with their subfunction. And we can then solve this by
taking a direct substitution approach to get one minus zero, which is of course
equal to one.

Now, here we might notice that the
third part of our question is very similar, instead asking us for the right-sided
limit as π₯ approaches zero of π of π₯.

When π₯ is greater than zero, π of
π₯ is defined by the third subfunction, one plus two π₯. We can evaluate this one-sided
limit in the same way, in putting our subfunction as π of π₯ and, again, taking a
direct substitution approach of π₯ equals zero to find that our limit is equal to
one. We have now answered part two and
three of the question, finding that both of our one-sided limits as π₯ approach zero
are equal to one.

Finally, part four of the question
asks, what type of discontinuity does the function π have at π₯ equals zero?

For this part of the question, we
first restate that both of the one-sided limits that we found exist are finite and
are equal to each other. Putting these two bits of
information together, we are also able to conclude that the normal limit as π₯
approaches zero also exists and is finite taking the same value, one. Letβs now look back at our answer to
part one. We found that when π₯ equals zero,
π is also zero. In other words, π of zero equals
zero. Again, letβs combine our
information. We have found that the limit as π₯
approaches zero of π of π₯ exists and is finite but is not equal to π of zero.

We now recall that this is the
exact condition that must be satisfied for a removal discontinuity at π₯ equals
zero. With this, we have answered all
four parts of our question. When π₯ equals zero, we have
evaluated our function, found its limits, and classified the type of discontinuity
that occurs. As a final note, if we were to
graph our function, it might look a little bit like this. And we could get rid of our removal
discontinuity by redefining π of zero equals one.

Okay, weβve just seen a piecewise
function, but in many cases our function will not be defined this way. Another thing we should be on the
lookout for is rational functions or functions with quotients of the form π of π₯
is equal to π of π₯ over π of π₯. In particular, we should pay
attention to values of π₯, which will make our denominator here π of π₯ equal to
zero. At these values of π₯, we would be
dividing by zero. And hence, we know that our function
π of π₯ would be undefined. Itβs worth noting that when we see
that a denominator of zero is possible at some value of π₯, we cannot immediately
conclude the type of discontinuity that will exist here. We can look back at two of the
examples that we saw earlier to illustrate this.

For π one, the denominator of the
quotient will be zero when π₯ equals three. For π four, the denominator of the
quotient will be zero when π₯ equals zero. Although in both cases we observe a
denominator of zero, the first is a removal discontinuity and the second is an
essential, an infinite discontinuity. In order to algebraically
distinguish between the two different types, we should still be examining our limits
as outlined by the criteria earlier. Now another point, in both of these
cases, it is very easy to find the value of π₯ which will make our denominator
zero. In some other cases, however, it
might not be immediately obvious.

Weβre still considering π of π₯ is
equal to π of π₯ over π of π₯. But here, letβs imagine that π of
π₯ is some polynomial. One tool that we can use to help
find that π of π₯ equals zero is the factor theorem. This tells us that if π₯ minus π is
a factor of π of π₯, then π of π must be zero. As a quick example, if π of π₯ was
some quadratic function and we were able to factorise it as π₯ minus π multiplied
by π₯ minus π, this means that the denominator of our function π of π₯ would be
zero when π₯ equals π and π₯ equals π. And hence, we should be evaluating
our function π for discontinuities at these values of π₯. Letβs see an example.

Consider the function π of π₯ is
equal to π₯ minus five divided by π₯ squared minus three π₯ minus 10. Find all the values of π₯ at which
π has discontinuities. Find the type of each
discontinuity.

To begin this question, we note
that π of π₯ is a rational function in the form π of π₯ over π of π₯. When seeing this form, we might
recall that our next step is to look for values of π₯, which would make our
denominator here called π of π₯ equal to zero. One tool that we can use to help us
find these values is the factor theorem. This tells us that if π₯ minus π
is a factor of π of π₯, then π of π must be equal to zero. Our first step should then be to
try and factorise our denominator. With a bit of inspection, we can
factorise our denominator to π₯ plus two times π₯ minus five. Here, the factor theorem tells us
that when π₯ equals negative two or when π₯ equals five, our denominator would be
zero. And hence π of π₯ would be
undefined.

We can extend our reasoning a
little further to say that these are the values at which weβll find our
discontinuities. But weβll need to look a little
closer to find their type. Letβs begin with π₯ equals negative
two. If we were to take the limit as π₯
approaches negative two π of π₯, using the factorised form of our denominator, and
we were to take a direct substitution approach, of course, we would find that our
limit does not exist, since we already know that we have a divide by zero. Again here, finding that a limit is
equal to a positive or negative infinity is a particular way of expressing that
limit doesnβt exist. From this information alone, we
could conclude that either the left, the right, or both of the one-sided limits also
do not exist.

However, letβs take a closer look
at them to see whatβs going on. We take a similar approach here,
but we noticed that since both of these parentheses are equal and nonzero, we can
cancel them out. We also noticed that, in the
denominator of our quotient, we have a negative two which weβre approaching from the
left and add two. This means that we have a one divided
by zero. But weβre approaching the zero from
the left. Since weβre approaching zero from the
left, we have a negative number. And if we divide one by an
infinitesimally small negative number, we get a negative infinity. This is a loose way in which we can
understand whether our infinity is a positive or a negative. We could follow this same process
to find that our right-sided limit would be a positive infinity.

Although these last two steps were
not strictly necessary, it does give us a better understanding as to what our
function will do as π₯ approaches negative two from the left and from the right,
respectively. Here, we have found that both our
left- and our right-sided limit as π₯ approaches negative two does not exist. And hence, at this point, we have
an essential discontinuity. And given our working, we could
more specifically classify this as an infinite discontinuity. Let us now move on to the case
where π₯ equals five. We evaluate the limit as π₯
approaches five of π of π₯. This time, direct substitution
leads us to the indeterminate form of zero over zero. And weβll be using a different
trick to get around this.

So far, weβve almost been ignoring
the fact that our original function π of π₯ appears to have a common factor of π₯
minus five on the top and bottom half of the quotient. If we were to cancel this factor, we
would be left with one over π₯ plus two. Now here, we must be very careful not
to say this is π of π₯. And instead, weβll call it something
different, say π of π₯. Although it seems like π of π₯ and
π of π₯ are equal, this is only true when π₯ is not equal to five, since we know
that π₯ equals five is not in the domain of π of π₯. Yet it is in the domain of π of
π₯. Herein, lies our trick. Since the limit concerns values of
π₯ which are arbitrarily close to five but not where π₯ is actually equal to five,
weβre able to say that the limit as π₯ approaches five of π of π₯ is equal to the
limit as π₯ approaches five π of π₯.

This now allows us to use a direct
substitution approach since π₯ equals five is in the domain of π of π₯. Doing so, we find that the value of
our limit is one over seven. Weβre now in a position to think
about our discontinuity. We found that the limit as π₯
approaches five of π of π₯ exists and is finite, taking the value of one over
seven. However, we concluded earlier that
when π₯ equals negative two or when π₯ equals five, π of π₯ was undefined since we
had a divide by zero. Hence, π of five is undefined. Putting these two bits of information
together, we have found that the limit as π₯ approaches five of π of π₯ exists, is
finite, but is not equal to π of five. This is the exact condition for a
removable discontinuity at π₯ equals five. We have now fully answered our
question.

We have found that π has an
essential and infinite discontinuity at π₯ equals negative two. And π has a removal discontinuity
at π₯ equals five. If we were to sketch the graph of
π of π₯, it might look a little like this. To finish off our video. Letβs go through some key
points. When a function does not satisfy
the continuity condition, it is said to be discontinuous. For discontinuities occurring at a
point, we have learned to classify them as the following, firstly, removable
discontinuities; secondly, essential discontinuities, which confer to be
subcategorized into infinite and oscillating discontinuities; and finally, jump
discontinuities. Each of the different types of
discontinuity has their own set of conditions outlined here.

When looking for discontinuities,
it is useful to evaluate certain points of a function. For piecewise defined functions,
look at values of π₯ at the boundary between the different subfunctions. And for quotients such as rational
functions in the form of π of π₯ over π of π₯, look for values of π₯ where the
denominator, π of π₯, equals zero. As a final note, in some cases
where a function is defined using absolute values or noninteger powers of π₯,
sometimes reexpressing this function into a more manageable form, which could
perhaps be piecewise, might make discontinuities easier to find and to classify.