### Video Transcript

Discuss the continuity of the
function π, given that π of π₯ is equal to the sin of negative eight π₯ plus 40
divided by π₯ minus five if π₯ is less than five and π of π₯ is equal to negative
eight π₯ squared divided by 25 if π₯ is greater than or equal to five.

The question wants us to decide on
which values of π₯ is the piecewise-defined function π of π₯ continuous. And we can find the values of π₯
for which a piecewise-defined function π of π₯ is continuous by following the
following three steps. First, we want to find the domain
of our piecewise-defined function π of π₯ since a function can only be continuous
at values of π₯ where itβs defined. Second, we want to check the
continuity of our function π of π₯ on each interval. Third, we just need to check that
the endpoints of our intervals match. In other words, weβre just
splitting the function π of π₯ into its individual pieces and then checking at the
end that they match up.

So, letβs start by finding the
domain of our function π of π₯. From the piecewise definition of
our function π of π₯, we can see when π₯ is less than five, our function π of π₯
is equal to the sin of negative eight π₯ plus 40 divided by π₯ minus five. And this is the quotient of a
trigonometric function and a polynomial, which are defined for all real values of
π₯. So, this quotient will be defined
as long as the denominator is not equal to zero. In other words, its domain is when
π₯ is not equal to five.

However, we can see we only use
values of π₯ less than five; π₯ is never equal to five. This tells us the sin of negative
eight π₯ plus 40 divided by π₯ minus five is defined for all of these values of
π₯. And in turn, this means our
function π of π₯ is defined for all values of π₯ less than five.

We can do the same for our second
interval. When π₯ is greater than or equal to
five, our function π of π₯ is equal to negative eight π₯ squared divided by 25. And we can see that this is a
polynomial, so itβs defined for all real values of π₯. So, we have negative eight π₯
squared divided by 25 is defined for all real values of π₯. In particular, itβs defined for all
values of π₯ greater than or equal to five. So, weβve shown π of π₯ is defined
for all values of π₯ less than five and for all values of π₯ greater than or equal
to five. In other words, the domain of our
function π of π₯ is the entire set of real numbers.

So, we found the domain of our
function π of π₯. We now need to check the continuity
of π of π₯ on each of the intervals. And to check the continuity on each
of these intervals, we recall the following three facts. Trigonometric functions are
continuous on their domain, polynomials are continuous on the entire set of real
numbers, and the quotient of two continuous functions is continuous on its
domain.

Letβs start when π₯ is less than
five. We see that our function π of π₯
is equal to the sin of negative eight π₯ plus 40 divided by π₯ minus five. This is the quotient of a
trigonometric function and a polynomial. So, by our rules, this function is
continuous on its domain. And we already found its
domain. Its domain is all the real values
of π₯ except for five. So, if π of π₯ is equal to the sin
of negative eight π₯ plus 40 divided by π₯ minus five when π₯ is less than five and
this is continuous, then π of π₯ must also be continuous on this interval.

We can do something similar for our
second interval, when π₯ is greater than or equal to five. We have that our function π of π₯
is the polynomial negative eight π₯ squared divided by 25. And polynomials are continuous for
all real values of π₯. So, when π₯ is greater than or
equal to five, our function π of π₯ is a polynomial, so itβs continuous on this
interval. However, we have to be careful
about when π₯ is equal to five. Remember, we need to check that the
endpoints of our intervals match. So, in fact, this only shows us
that our function π of π₯ is continuous when π₯ is greater than five. Weβll need to check the case for π₯
is equal to five directly.

Letβs start with the endpoint of
our interval when π₯ is less than five. Since our interval has π₯ strictly
less than five, weβll take the limit as π₯ approaches five from the left of the sin
of negative eight π₯ plus 40 divided by π₯ minus five. This is the quotient of a
trigonometric function and a linear function, so we can try using direct
substitution. Substituting π₯ is equal to five,
we get the sin of negative eight times five plus 40 divided by five minus five. And if we evaluate this expression,
we get the sin of zero divided by zero, which is the indeterminate form zero divided
by zero. So, weβre going to need to try and
evaluate this limit in a different way.

To help us evaluate this limit, we
recall the following standard limit law. The limit as π₯ approaches zero of
the sin of π₯ divided by π₯ is equal to one. At first, it might seem difficult
to see how we can use this result. However, we can notice that
negative eight π₯ plus 40 is actually equal to negative eight times π₯ minus
five.

And now, we can see something
interesting; our limit has π₯ approaching five from the left. This means our denominator of π₯
minus five is approaching zero. And the factor of π₯ minus five in
our numerator is also approaching zero. This is very similar to our limit
law. This is motivation to rewrite our
limit by using the substitution π’ is equal to π₯ minus five.

Inside of our limit, by using this
substitution, we get the sin of negative eight π’ all divided by π’. However, in our original limit, π₯
was approaching five from the left. In other words, π₯ is getting
closer and closer to five while our values of π₯ are less than five. We can ask the question: what does
this mean for our values of π’? Well, π’ is equal to π₯ minus
five. And π₯ is getting closer and closer
to five but always being less than five. This means π’ is approaching zero
and π’ is less than zero. In other words, as π₯ approaches
five from the left, π’ approaches zero from the left.

So, now, we have the limit as π’
approaches zero from the left of the sin of negative eight π’ divided by π’. And this is now very similar to our
limit law, except we need to deal with the coefficient of negative π₯ in front of
the π₯ inside of our sine function. Weβll start by rewriting all
instances of π₯ with negative eight π’. Next, we can take the constant
factor of negative one-eighth outside of our limit. And if negative eight π’ is
approaching zero, this is the same as saying that π’ is approaching zero.

Finally, weβll just multiply
through by negative eight, giving us the limit as π’ approaches zero of the sin of
negative eight π’ divided by π’ is equal to negative eight. And we can see this is exactly
equal to what we have in our limit.

So, we found the first endpoint of
our interval. Itβs negative eight. Itβs much easier to find the second
endpoint of our interval. Since our interval is π₯ is greater
than or equal to five, the endpoint will just be when π₯ is equal to five. So, we just substitute π₯ is equal
to five into negative eight π₯ squared divided by 25. This gives us negative eight times
five squared divided by 25, which we can evaluate to give us negative eight.

So, we can see, in both cases, we
get negative eight. So, our function π of π₯ is also
continuous when π₯ is equal to five. So, weβve shown our function π of
π₯ is continuous when π₯ is less than five, when π₯ is greater than five, and when
π₯ is equal to five. Therefore, weβve shown that the
function π is continuous for all real values of π₯.