### Video Transcript

Is the function π of π₯, which is
piecewise defined to be two π₯ plus four all over π₯ plus two if π₯ is less than
negative two. Zero if π₯ is equal to negative
two. And π₯ squared plus six π₯ plus
eight all over π₯ plus two if π₯ is greater than negative two, continuous at π₯
equals negative two.

To find out, we go through our
checklist. For the function π to be
continuous at π₯ equals negative two, π of negative two must be defined. The limit of π of π₯ as π₯
approaches negative two must exist. And these values must be equal. So we start by checking that π of
negative two is defined. Well, thatβs easy to see from the
question. Weβre told that π of π₯ is zero if
π₯ is negative two. So π of negative two is
defined. π of negative two is equal to
zero. Now, we need to check that the
limit of π of π₯ as π₯ approaches negative two exists. There are different rules for the
values of π of π₯ when π₯ is less than negative two and when π₯ is greater than
negative two. Which weβll use to check the
left-hand and right-hand limits, making sure that they exist and are equal.

Whatβs the left-hand limit? Well, to the left of negative two,
π of π₯ is defined by two π₯ plus four all over π₯ plus two. So we have to find the limit of two
π₯ plus four all over π₯ plus two, as π₯ approaches negative two from the left. If we try direct substitution, we
get the indeterminate form zero over zero. So we have to find this limit some
other way. We do this by factoring the
numerator, getting two times π₯ plus two. And this factor of π₯ plus two
cancels with the factor of π₯ plus two in the denominator. And so our limit is that of the
constant function two whose value must be two. So the left-hand limit exists. What about the right-hand one?

To the right of π₯ equals negative
two, π of π₯ is this algebraic fraction. And we can find the value of this
limit in a similar way. Factoring the numerator and
canceling the common factor of π₯ plus two, to get the limit of π₯ plus four as π₯
approaches negative two from the right. This can be solved using direct
substitution. We get negative two plus four,
which is two. So this right-hand limit also
exists and is equal to the left-hand limit. Both of them have the value
two. Hence the limit of π of π₯ as π₯
approaches negative two from either direction also exists and has the value two.

All thatβs left to check is this
third point. For π of π₯ to be continuous at π₯
equals negative two, we need the limit of π of π₯ as π₯ approaches negative two to
be equal to the value of π at negative two, π of negative two. But weβve just seen that the value
of this limit is two, whereas π evaluated at negative two is zero. The limit of π of π₯ as π₯
approaches negative two is not equal to π of negative two. Although the first two conditions
on our checklist are satisfied, the last one isnβt. And so our answer is no. The function π is not continuous
at π₯ equals negative two, as the limit of π of π₯ as π₯ approaches negative two is
not equal to π of negative two.