### Video Transcript

Suppose that π of π₯ is equal to
five sin of π₯ minus three over π₯ minus three if π₯ is less than three and five π₯
squared over nine if π₯ is greater than or equal to three. Find the set on which π is
continuous.

Here, we have a piecewise function
defined over two intervals. The boundary of our two intervals
occurs at the point where π₯ equals three. And hence, this is an important
point. For our first subfunction, we have
a trigonometric expression on the top-half of our quotient and a binomial on the
bottom-half. Since we know that trigonometric
functions and polynomials are continuous on their domains, and quotients of
continuous functions are also continuous on their domains, we conclude that this
subfunction is, therefore, also continuous on its domain.

Now, here, we need to be slightly
careful since, at values where π₯ equals three, this subfunction evaluates to zero
over zero, which is an indeterminate form. Lucky for us, π of π₯ is only
defined by this subfunction at values of π₯ which are strictly less than three, not
where π₯ is equal to three. Instead, when π₯ is greater than or
equal to three, π of π₯ is defined by five π₯ squared over nine.

Again, here, itβs worth noting that
this monomial is defined over and has a domain of all real numbers. This means that the domain of our
function π of π₯ is all the real numbers. But we should be careful not to
hastily jump to the conclusion that π of π₯ is also continuous over all the real
numbers. Instead, we must still check the
criteria for continuity at the boundary between our subfunctions, or when π₯ is
equal to three.

First, letβs find π of three by
substituting in to five π₯ squared over nine. This value is easy to compute. And we get an answer of five. Next, we need to check that our
normal limit exists and is also equal to five. If this is not the case, then weβll
have a discontinuity at π₯ equals three. And hence, our function will not be
continuous at this point.

To move forward, we first recognise
that, either side of π₯ equals three, our function is defined by two different
subfunctions. To find our left limit, or when π₯
approaches from the negative direction, we use our first subfunction. Here, weβve already shown that a
direct substitution of π₯ equals three leads us to an indeterminate form of zero
over zero, so weβll need to use a different approach. Instead, we use the rule that the
limit, as π₯ approaches zero of sin π₯ over π₯, is equal to one.

Now, our expression isnβt in this
form. And so, we perform some
manipulations first by taking a factor of five outside of our limit using the
constant multiple rule. Next, we perform a
π’-substitution. By setting π’ as π₯ minus three, we
get the following. Our limit becomes sin π’ over
π’. However, we mustnβt forget to
change the limit value itself. π’ plus three is equal to π₯. Hence, π’ plus three approaches
three from the negative direction, or π’ approaches zero from the negative
direction.

Looking at our rule, we know that
if the normal limit exists and is equal to one, then the left- and right-sided
limits also exist and are also equal to one. We can now use our rule to evaluate
that this limit is equal to one. Hence, the left-sided limit, as π₯
approaches three of π of π₯, is equal to five times one, which is, of course, just
five. Now, our right-sided limit is far
easier to evaluate. For this, since weβre approaching
π₯ equals three from the right, we take the limit using our other subfunction. Simply, by direct substitution, we
find that this limit is equal to five.

Since the left- and right-sided
limits both exist and are equal to the same value, we can, therefore, conclude that
the limit, as π₯ approaches three of π of π₯, is also equal to five. And earlier, youβll recall that we
found that π of three is also equal to five. Since the limit, as π₯ approaches
three of π of π₯, is equal to π of three, we conclude that π of π₯ is continuous
when π₯ is equal to three.

If you were to think about this
visually, this means that the two end points of our subfunctions will join up. Let us now recall that, earlier, we
concluded that π of π₯ was continuous over all of the real numbers aside from
three, which we had to check. And now that we have checked three,
weβre in a position to say that our function π of π₯ is continuous on all the real
numbers. And this is the answer to our
question.