### Video Transcript

Find all the values of π for which
the limit of the function exists as π₯ approaches zero when π of π₯ is equal to
negative six π minus 18 times the cos of π₯, when π₯ is less than zero, and π of
π₯ is equal to the tangent of 12π₯ divided by π times π₯, when π₯ is greater than
zero.

In this question, weβre given a
piecewise-defined function π of π₯, which contains an unknown value of π. We need to determine all of the
possible values of π for which the limit of this function exists as π₯ is
approaching zero. To answer this question, we should
start by recalling what it means for the limit of a function at a point to
exist. However, thereβs many different
ways of writing this and thereβs many different ways for a limit to not exist. So instead, itβs usually a good
idea to take a look at the functionβs outputs around this value of π₯. This can give a useful idea about
what might happen to the function.

And we can notice something very
interesting. π₯ is equal to zero is one of the
endpoints of the subdomains of this function. This means our function has a
different definition when π₯ is less than zero and a different definition when π₯ is
greater than zero. So for this limit to exist, weβre
going to need to analyze its left and right limits. We can then recall that we say the
limit as π₯ approaches π of a function π of π₯ exists and is equal to a finite
value of π if the limit as π₯ approaches π from the right of π of π₯ and the
limit as π₯ approaches π from the left of π of π₯ both exist and are both equal to
π.

This means we can check that the
limit of a function exists by checking four things. First, we need to check that the
limit from the right exists. Second, we need to check that this
is equal to π. Third, we need to check that the
limit from the left exists. And finally, we need to check that
this is also equal to π.

To apply this to our function,
weβre going to need to evaluate its left and right limits. Letβs start with the left
limit. Since our value of π is zero, this
is the limit as π₯ approaches zero from the left of π of π₯. And since weβre taking the limit as
π₯ approaches zero from the left of π of π₯, our values of π₯ are less than
zero. So our function π of π₯ is exactly
equal to negative six π minus 18 cos of π₯. And if the functions are equal,
then their limits as π₯ approaches zero from the left will be equal. We just need to find the limit as
π₯ approaches zero from the left of negative six π minus 18 cos of π₯. And this is a constant plus a
trigonometric function, so we can just do this by using direct substitution.

Substituting π₯ is equal to zero
into this function, we get negative six π minus 18 multiplied by the cos of
zero. And the cos of zero is one, so this
is just equal to negative six π minus 18. And itβs worth noting this is a
finite value for any real value of π. So weβve shown that this limit
exists and we found our value of π. So letβs keep this in mind while we
try to evaluate the limit as π₯ approaches zero from the right of π of π₯. This time, since weβre taking the
limit as π₯ approaches zero from the right, our values of π₯ are going to be greater
than zero, which means our function π of π₯ is exactly equal to the tan of 12π₯
divided by π times π₯. And since the functions are equal
to the right of π₯, the limit as π₯ approaches zero from the right will also be
equal.

This means we just need to
determine the limit as π₯ approaches zero from the right of the tan of 12π₯ divided
by π times π₯. And this limit is incredibly
similar to one of our trigonometric limit results. We know for any real constant π,
the limit as π₯ approaches zero of the tan of ππ₯ divided by π₯ is equal to π. And itβs worth noting these results
will hold true as left and right limits since the left limit and the right limit
need to be equal for the standard limit to exist. We canβt directly apply this,
however, since weβre dividing by π times π₯ in our denominator.

So we need to take a factor of one
over π outside of our limit. This then gives us one over π
multiplied by the limit as π₯ approaches zero from the right of the tan of 12π₯
divided by π₯. And by our limit result this is
equal to the coefficient of π₯, which is 12. So we get 12 times one over π,
which is 12 over π. Therefore, the limit as π₯
approaches π from the right of π of π₯ exists, provided π is not equal to
zero. And we already knew that π was
nonzero since weβre dividing by π in our definition of π of π₯.

And this means we have one final
thing we need to check. We need the limit as π₯ approaches
zero from the left of π of π₯ to be equal to the limit as π₯ approaches zero from
the right of π of π₯ for this limit to exist. We can substitute the expressions
we found for each of these limits. We need negative six π minus 18 to
be equal to 12 divided by π. To solve this equation, weβll start
by multiplying both sides of the equation through by π and dividing through by
six. This gives us that negative π
squared minus three π is equal to two. We can then rearrange this
quadratic to get π squared plus three π plus two is equal to zero and then factor
the quadratic fully. We get π plus one multiplied by π
plus two is equal to zero.

And of course, for a product to be
equal to zero, one of the factors must be zero. So we either have π is negative
one or π is negative two. And we can write these as a set,
the set containing negative one and negative two, which is our final answer. Therefore, we were able to show
there are only two values for which the limit as π₯ approaches zero of π of π₯ is
equal to negative six π minus 18 cos of π₯ when π₯ is less than zero and π of is
equal to tan of 12π₯ divided by ππ₯ when π₯ is greater than zero exists and must be
in the set containing negative one and negative two.