### Video Transcript

Determine the limit as π₯
approaches zero of negative four tan cubed of π₯ plus one all raised to the power of
cot cubed of π₯.

In this question, weβre asked to
evaluate a limit. And the first thing we should do
whenever weβre asked to evaluate a limit is see if we can do this directly. We might want to try direct
substitution. However, remember, trigonometric
functions are only continuous on their domain. And π₯ is equal to zero is not in
the domain of cot of π₯. So we canβt evaluate this by using
direct substitution.

However, we can see what happens to
our output as π₯ is getting closer to zero. Inside our parentheses, as the
values of π₯ are getting closer to zero, the tan of π₯ is also getting closer to
zero. So the expression inside our
parentheses is getting closer to one. However, we canβt determine what
happens to our exponent as π₯ approaches zero. As π₯ approaches zero from the
right, cot of π₯ approaches β. However, as π₯ approaches zero from
the left, cot of π₯ approaches negative β. And in either case, one to the
power of positive β and one to the power of negative β are indeterminate forms
anyway. So we canβt use this to evaluate
our limit.

Instead, weβre going to need to try
some form of manipulation. And it can be difficult to see what
we need to do to help us evaluate this limit. One thing we can notice is cot
cubed of π₯ is actually one divided by tan cubed of π₯. So weβll try using this to help us
evaluate our limit. We recall this tells us the limit
as π approaches zero of one plus π all raised to the power of one over π is equal
to π. And itβs worth pointing out here we
can use our other limit result involving Eulerβs constant π to answer this
question.

However, usually, one of the two
limit results involves a lot easier working. And it can be very difficult to
determine which of our two limit results we should use just by looking at the limit
weβre asked to evaluate. So if weβre struggling, we should
try switching limit results.

Now, to use this limit to help us
evaluate this question, weβre going to need to start by rewriting our limit in this
form. The first thing weβll do is switch
the order of the two terms inside our parentheses. This gives us the limit as π₯
approaches zero of one minus four tan cubed of π₯ all raised to the power of cot
cubed of π₯. Next, inside our parentheses, weβre
going to want one plus our variable. So weβre going to need to use a
substitution. Weβre going to substitute π is
equal to negative four times the tan cubed of π₯.

So to rewrite our limit by using
the substitution, we need to see what happens to our values of π as π₯ approaches
zero. We can do this directly from our
substitution. On the right-hand side of our
equation, as the values of π₯ approach zero, the right-hand side of our equation is
approaching zero, because the right-hand side of this equation is a continuous
function. So we can actually just evaluate
this by direct substitution. So as our values of π₯ are
approaching zero, our values of π are also approaching zero.

But thereβs one more instance of π₯
which appears in our limit. We need to find an expression for
cot cubed of π₯ in terms of π. And to do this, weβre going to need
to rearrange our substitution. We start by dividing through by
negative four. We get negative π over four is
equal to tan cubed of π₯.

And now, because we want an
expression for cot cubed of π₯, we can actually just take the reciprocal of both
sides of this equation. And itβs worth pointing out here we
donβt need to worry about dividing by zero because our limits are as π₯ approaches
zero and π approaches zero. So our values of π₯ and π are
never equal to zero. And this means tan cubed of π₯ is
not equal to zero and π is not equal to zero. So this gives us that negative four
over π is equal to cot cubed of π₯.

Now weβre ready to use this
substitution to rewrite our limit. Weβve rewritten our limit as the
limit as π approaches zero of one plus π all raised to the power of negative four
over π. And now this is almost in exactly
our form of using our limit result involving Eulerβs constant π. The only difference is, in our
exponent, we have a factor of negative four. So to apply our limit result, weβre
going to want to take this exponent factor of negative four outside of our
limit. To do this, weβre going to need to
use the power rule for limits. But first, weβre going to use our
laws of exponents.

We know π to the power of π times
π is equal to π to the power of π all raised to the power of π. By setting π equal to one plus π,
π equal to one over π, and π equal to negative four, we can rewrite our limit as
the limit as π approaches zero of one plus π all raised to the power of one over
π all raised to the power of negative four. Now, all thatβs left to do is take
this exponent of negative four outside of our limit. And to do this, weβre going to need
to use the power rule for limits.

One version of this tells us the
limit as π approaches π of π of π raised to the πth power is equal to the limit
as π approaches π of π of π all raised to the πth power, provided the limit as
π approaches π of π of π exists and we can raise this to the πth power. And the easiest way to show that
these two prerequisites are true is to write the next step in our line of
working. By using the power rule for limits,
we get that our limit is equal to the limit as π approaches zero of one plus π all
raised to the power of one over π all raised to the power of negative four.

But remember, this is provided the
limit as π approaches zero of one plus π all raised to the power of one over π
exists, and we know this does exist. Itβs our limit result involving
Eulerβs constant π. And of course, donβt forget we also
need to check that we can raise this number to our exponent, and of course we
can. We just get π to the power of
negative four. This means we were able to justify
our use of the power rule for limits to get that our limit evaluates to π to the
power of negative four. However, weβll do one last piece of
manipulation to rewrite this as one over π to the fourth power, which is our final
answer.

Therefore, we were able to show the
limit as π₯ approaches zero of negative four tan cubed of π₯ plus one all raised to
the power of cot cubed of π₯ is equal to one over π to the fourth power.