### Video Transcript

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.