### Video Transcript

Discuss the continuity of the
function π at π₯ is equal to π by two, given that the function π of π₯ is equal
to eight plus the cos of seven π₯ if π₯ is less than π by two, and the function π
of π₯ is equal to seven plus the sin of five π₯ if π₯ is greater than or equal to π
by two.

The question gives us a
piecewise-defined function π of π₯ and asks us to decide if the function is
continuous at the value π₯ is equal to π by two. Letβs start by recalling what it
means for a function to be continuous at a point. We say that a function π of π₯ is
continuous at the value π₯ is equal to π if the following three conditions are
true.

First, π evaluated at π must be
defined. This is the same as saying that π
is in the domain of our function π. Second, we must have the limit as
π₯ approaches π of π of π₯ exists. And itβs worth noting this is the
same as saying that both the limit as π₯ approaches π from the left of π of π₯ and
the limit as π₯ approaches π from the right of π of π₯ exist and are equal. Finally, we need the limit as π₯
approaches π of π of π₯ to be equal to π evaluated at π. Weβre only interested in the
continuity of π when π₯ is equal to π by two, so weβll set π equal to π by
two. So thatβs right. π is π by two in our definition
for continuity.

The first thing we need to check is
π is defined at π₯ is equal to π by two. And we can do this directly from
the piecewise definition of π of π₯. We can see when π₯ is equal to π
by two, our function π of π₯ is equal to seven plus the sin of five π₯. So π evaluated at π by two is
equal to seven plus the sin of five times π by two. And we can evaluate this by using
the fact that the sine function is periodic about an interval of two π. This tells us that sin of five π
by two is equal to the sin of π by two which we know is equal to one. So this tells us π evaluated at π
by two is equal to seven plus one, which is eight. So weβve shown that π evaluated at
π by two is defined. In other words, π by two is in the
domain of our function π of π₯.

Letβs move on to our second
condition for continuity. We need to show the limit as π₯
approaches π by two of π of π₯ exists. And weβll do this by showing the
limit as π₯ approaches π by two from the left of π of π₯ is equal to the limit as
π₯ approaches π by two from the right of π of π₯. And of course, showing that both of
these limits exist. Letβs start by finding the limit as
π₯ approaches π by two from the left of π of π₯. Since π₯ is approaching π by two
from the left, we must have our values of π₯ are less than π by two. And we can see from our piecewise
definition of the function π of π₯, when our values of π₯ are less than π by two,
our function π of π₯ is exactly equal to eight plus the cos of seven π₯.

And if these functions are exactly
equal for values of π₯ less than π by two, their limits as π₯ approaches π by two
from the left will also be equal. So we want to evaluate the limit as
π₯ approaches π by two from the left of eight plus the cos of seven π₯. But this is just a constant plus a
trigonometric function. We can do this by direct
substitution. Substituting π₯ is equal to π by
two, we get eight plus the cos of seven π by two. And again, the cos function is
periodic about an interval of two π. So the cos of seven π by two is
equal to the cos of three π by two, which we know is equal to zero. So weβve shown this limit evaluates
to give us eight plus zero, which is equal to eight. So the limit as π₯ approached π by
two from the left of π of π₯ was equal to eight.

We now need to show that our limit
as π₯ approaches π by two from the right of π of π₯ is also equal to eight. And we can actually check this in
exactly the same way. Weβll take the limit as π₯
approaches π by two from the right of π of π₯. This means that our values of π₯
are greater than π by two. We then see from our piecewise
definition of the function π of π₯, if our values of π₯ are greater than π by two,
our function π of π₯ is exactly equal to seven plus the sin of five π₯. So their limits as π₯ approaches π
by two from the right will be equal.

And now we see we want to evaluate
the limit of a constant plus a trigonometric function. We can do this by direct
substitution. Substituting π₯ is equal to π by
two, we get seven plus the sin of five π by two. And we actually already evaluated
this expression. Itβs actually equal to π evaluated
at π by two. So we can just say this is equal to
eight. So weβve shown the limit as π₯
approaches π by two from the right of π of π₯ is also equal to eight. So both of these limits exist and
theyβre equal. So our second continuity condition
is also true.

Finally, we just need to check our
last continuity condition. The limit as π₯ approaches π by
two of π of π₯ needs to be equal to π evaluated at π by two. We actually evaluated both of these
expressions in the first and second part of our continuity condition. We evaluated π at π by two in the
first part of our continuity condition. We showed that it was equal to
eight.

And itβs worth reiterating at this
point when we did our second continuity condition and we showed that the limit from
the left and the limit from the right of π of π₯ were both equal. In this case, we showed that both
of these were equal to eight. This is actually equivalent to just
showing the limit as π₯ approaches π by two of π of π₯ is equal to eight. So weβve already shown that both of
these expressions are equal to eight. So our third continuity condition
is also true.

Therefore, since weβve shown all
three of our continuity conditions are true for the function π of π₯ at π₯ is equal
to π by two. We can conclude that the function
is continuous at π₯ is equal to π by two.