### Video Transcript

Discuss the continuity of the
function π at π₯ is equal to zero, given that π of π₯ is equal to six π₯ squared
minus the sin squared of π₯ divided by π₯ times the tan of four π₯ when π₯ is less
than zero and π of π₯ is equal to π₯ plus five divided by π₯ plus four when π₯ is
greater than or equal to zero.

In this question, weβre given a
piecewise-defined function π of π₯ and we want to determine the continuity of the
function π when π₯ is equal to zero. To answer this question, letβs
start by recalling the three conditions which need to hold for a function to be
continuous at a point. For a function π to be continuous
at a value of π₯ is equal to π, we need π evaluated at π to be defined. This means π needs to be in the
domain of our function π. We need the limit as π₯ approaches
π of π of π₯ to exist. And we also need this limit as π₯
approaches π of π of π₯ to be equal to π evaluated at π. If all three conditions hold, then
we say π is continuous at π₯ is equal to π.

In this question, weβre using the
piecewise-defined function π. And we want to determine its
continuity when π₯ is equal to zero. So we can substitute π is equal to
zero into our definition. We now need to check all three of
these conditions. Letβs start with the first
condition. We want to check that zero is in
the domain of our function π. And since this is a
piecewise-defined function, we can see this directly from its definition. Zero is included in the second
subdomain. And although we could conclude zero
is in the domain of our function, itβs worth noting we need the value of π
evaluated at zero for the third condition. So weβll evaluate π at zero
now. We substitute π₯ is equal to zero
into our second subfunction. π evaluated at zero is zero plus
five divided by zero plus four, which evaluates to give us five over four. Therefore, the first condition of
continuity for π at zero is true.

Letβs now move on to the second
condition. We want to check that the limit as
π₯ approaches zero of π of π₯ exists. And the exact method we should use
to check that this function exists will depend entirely on the function weβre given
and the point. In this case, we can note weβre
given a piecewise-defined function and our value of π₯ is equal to zero is one of
the endpoints of the subdomain. This means our function changes
definition to the left of π₯ is equal to zero and to the right of π₯ is equal to
zero. This should hint to us that we can
check the limit as π₯ approaches zero of π of π₯ by evaluating its left and right
limits.

So letβs start by evaluating the
limit as π₯ approaches zero from the left of π of π₯. Since weβre taking the limit as π₯
approaches zero from the left, our values of π₯ are all going to be less than
zero. And we can see when π₯ is less than
zero, our function π of π₯ is exactly equal to its first subfunction. So their limits will be equal. We just need to evaluate the limit
as π₯ approaches zero from the left of six π₯ squared minus the sin squared of π₯
divided by π₯ times the tan of four π₯. And we might be tempted to do this
by direct substitution. However, this wonβt work. In our numerator, weβll get zero
minus zero, which is equal to zero. And our denominator also has a
factor of zero. So instead weβre going to need to
simplify this limit and then apply our trigonometric limit results.

Letβs start by dividing both terms
in the numerator separately by the denominator and splitting our limit over the
difference. This gives us the limit as π₯
approaches zero from the left of six π₯ squared divided by π₯ times the tan of four
π₯ minus the limit as π₯ approaches zero from the left of the sin squared of π₯
divided by π₯ multiplied by the tan of four π₯. And we can simplify this
slightly. In our first limit, we can cancel
the shared factor of π₯ in the numerator and denominator.

Weβre now almost ready to evaluate
these limits. Weβll start by recalling that the
limit as π₯ approaches zero of π₯ divided by the tan of ππ₯ is equal to one divided
by π provided our value of π is not equal to zero. This will allow us to evaluate the
first limit. We just need to take out the
constant factor of six. We can then see that we have π₯
divided by the tan of four π₯ inside of our limit. So our value of π is equal to
four. And of course, we need to multiply
this by six. So we get six multiply by
one-quarter. We now want to use this result
along with the fact that the limit as π₯ approaches zero of the sin of ππ₯ divided
by π₯ is equal to π for any real value of π to evaluate our second limit.

However, we canβt do this directly
because our limit is not quite in this form. Instead, weβre going to need to
multiply both the numerator and denominator of this limit by π₯. We can now see we have sin squared
of π₯ in our numerator and π₯ squared in our denominator. And we also have π₯ in our
numerator and tan of four π₯ in the denominator. This means we can use the productβs
rule for limits to rewrite this limit as the limit as π₯ approaches zero from the
left of π₯ divided by the tan of four π₯ multiplied by the limit as π₯ approaches
zero from the left of sin squared of π₯ divided by π₯ squared provided both of these
limits exist. And weβve already seen how to
evaluate this first limit by using our trigonometric limit result. Our value of π is four, so itβs
equal to one-quarter.

To evaluate our second limit
however, we need to notice we have a square in our numerator and a square in our
denominator. We can take the square outside of
our limit by using the power rule for limits. This then gives us the limit as π₯
approaches zero from the left of the sin of π₯ divided by π₯ all squared. And now, this is our trigonometric
limit result with the value of π equal to one. So the limit evaluates to give us
one, and one squared is equal to one. Therefore, weβve shown this is
equal to six times one-quarter minus one-quarter multiplied by one. And we can then just evaluate
this. Six-quarters minus one-quarter is
five-quarters. Therefore, the limit as π₯
approaches zero from the left of π of π₯ exists and itβs equal to five over
four.

Letβs now clear some space and
evaluate the limit as π₯ approaches zero from the right of π of π₯. This time, since weβre evaluating
the limit as π₯ approaches zero from the right of π of π₯, our values of π₯ are
going to be greater than zero. And we can see when π₯ is greater
than zero, our function π of π₯ is exactly equal to the rational function π₯ plus
five divided by π₯ plus four. Therefore, their limits as π₯
approaches zero from the right must be equal. So we just need to evaluate the
limit as π₯ approaches zero from the right of π₯ plus five divided by π₯ plus
four. And this is a rational
function. So we can attempt to do this by
using direct substitution. And we already found this when we
evaluated π at zero. Itβs equal to zero plus five
divided by zero plus four, which is equal to five-quarters.

Therefore, the limit as π₯
approaches zero from the right of π of π₯ exists, and itβs equal to five over
four. And the limit as π₯ approaches zero
from the left of π of π₯ exists, and itβs also equal to five over four. Since the left and right limit both
exist and are equal, we can conclude the limit as π₯ approaches zero of π of π₯
exists and itβs equal to five over four.

And for our third and final
condition, we just need to check that the limit as π₯ approaches zero of π of π₯ is
equal to π evaluated at zero. We just showed the left and right
limit of π of π₯ both exist and are both equal to five-quarters. So the limit as π₯ approaches zero
of π of π₯ is equal to five-quarters. And weβve also already calculated
π evaluated at zero; itβs also equal to five-quarters. Therefore, the third continuity
condition holds, and we can conclude the function π given to us in the question is
continuous at π₯ is equal to zero.