### Video Transcript

Determine the limit as π₯
approaches zero of negative six times the sin of five π₯ all divided by two to the
power of π₯ minus one.

The question is asking us to
determine the limit of the quotient of two functions. We can see the function in our
numerator is made up of polynomials, compositions, and standard trigonometric
functions. Similarly, the function in the
denominator is an exponential function, and then we just subtract the constant
one. So we could attempt to evaluate the
limit of our numerator and the limit of our denominator by direct substitution.

So by substituting π₯ is equal to
zero, we get negative six times the sin of five times zero all divided by two to the
zeroth power minus one. We know the sin of zero is equal to
zero, so our numerator is equal to zero. And two to the zeroth power is just
equal to one, so our denominator is also equal to zero. So using direct substitution gave
us an indeterminate form. This means weβre going to need to
evaluate our limit in a different method.

We recall if the limit of the
quotient of two functions gives us an indeterminate form, then we can try using
LβHΓ΄pitalβs rule. One version of LβHΓ΄pitalβs rule
tells us that if we have two differentiable functions π and π, where the
derivative of π is not equal to zero for values of π₯ around π, although π prime
of π is allowed to be equal to zero. And we also have the limit as π₯
approaches π of π of π₯ and the limit as π₯ approaches π of π of π₯ are both
equal to zero. Then the limit as π₯ approaches π
of π of π₯ divided by π of π₯ is equal to the limit as π₯ approaches π of π
prime of π₯ divided by π prime of π₯.

In other words, if the limit as π₯
approaches π of the quotient of two functions gives us an indeterminate form, then,
under certain circumstances, we can evaluate this limit as the limit as π₯
approaches π of the quotient of the derivatives of these two functions. So letβs see if we can apply this
version of LβHΓ΄pitalβs rule to the limit given to us in the question.

Weβll set our function π of π₯ to
be the function in the numerator of our limit. Thatβs negative six times the sin
of five π₯. And weβll set our function π of π₯
to be the function in the denominator of our limit. Thatβs two to the power of π₯ minus
one. Finally, since we want to find the
limit as π₯ approaches zero, weβll set our value of π equal to zero.

We now need to check that all of
the prerequisites for LβHΓ΄pitalβs rule are true. First, letβs check that our
functions π and π are differentiable. We already said that our numerator
is the composition of polynomials and standard trigonometric functions. Since all of these are
differentiable, their compositions will also be differentiable. So our numerator being the
composition of differentiable functions must be differentiable. Itβs a similar story in our
denominator. We know the exponential function
two to the power of π₯ is differentiable. We then subtract one from this and
itβs still differentiable.

Next, we need to show that the
derivative of π of π₯ is not equal to zero around π₯ is equal to zero. To check that this is true, letβs
find an expression for π prime of π₯. We recall that π of π₯ is equal to
two to the power of π₯ minus one. Thereβs a few different ways of
differentiating this. We could recall a standard result
to differentiate exponential functions. However, we can do this
directly.

We recall by using our laws of
exponents and our laws of logarithms π to the power of π₯ is equal to π to the
power of the natural logarithm of π all raised to the power of π₯. This is because π to the power of
the natural logarithm of π is just equal to π. Then, by using our laws of
exponents, we can rewrite this as π to the power of π₯ times the natural logarithm
of π. Using this, we can rewrite our
function π of π₯ as π to the power of π₯ times the natural logarithm of two minus
one.

We now see this is in a form which
we can differentiate. For any constant π, we know the
derivative of π to the power of ππ₯ with respect to π₯ is equal to π times π to
the power of ππ₯. So we can differentiate each term
in our function for π of π₯ to get π prime of π₯ is equal to the natural logarithm
of two times π to the power of π₯ times the natural logarithm of two minus
zero. And of course, we can just remove
the minus zero.

Remember, weβre checking that π
prime of π₯ is not equal to zero around π₯ is equal to zero. We can see that the natural
logarithm of two is a positive constant. And we have π raised to the power
of any number is also a positive constant. So we have that π prime of π₯ is
positive. So weβve shown that the derivative
function is positive. In particular, this means that itβs
not equal to zero for any value of π₯.

The last prerequisite we need to
show is true is that the limit as π₯ approaches zero of π of π₯ and the limit as π₯
approaches zero of π of π₯ need to both be equal to zero. However, weβve already shown this
to be true. When we initially try to evaluate
our limit by using direct substitution, the limit we evaluated in the numerator is
equal to the limit as π₯ approaches zero of π of π₯. We evaluated both of these limits
by using direct substitution. And we got zero divided by
zero. So in fact, weβve already shown
that both of these limits are already equal to zero. So now, weβve shown that all of our
prerequisites are true.

This means instead of evaluating
the limit given to us in the question directly, we can instead try to evaluate the
limit as π₯ approaches zero of π prime of π₯ divided by π prime of π₯. Before we do this, letβs first find
an expression for π prime of π₯. We recall π of π₯ is equal to
negative six times the sin of five π₯. And we can differentiate this by
using one of our derivative rules for trigonometric functions. For constants π and π, the
derivative of π times the sin of ππ₯ with respect to π₯ is equal to π times π
times the cos of ππ₯.

Applying this, we get that π prime
of π₯ is equal to five times negative six times the cos of five π₯, which we can
simplify to give us negative 30 times the cos of five π₯. So weβre now ready to attempt to
evaluate this limit by using LβHΓ΄pitalβs rule. First, this version of LβHΓ΄pitalβs
rule told us the limit as π₯ approaches zero of π of π₯ divided by π of π₯ is
equal to the limit as π₯ approaches zero of π prime of π₯ divided by π prime of
π₯.

Next, we showed that π prime of π₯
is equal to negative 30 times the cos of five π₯ and π prime of π₯ is equal to the
natural logarithm of two times π to the power of π₯ times the natural logarithm of
two. We can again see that both of these
functions are made up of standard functions. So we can attempt to evaluate this
by using direct substitution. Substituting π₯ is equal to zero,
we get negative 30 times the cos of five times zero all divided by the natural
logarithm of two times π to the power of zero times the natural logarithm of
two.

We know the cos of zero is just
equal to one and zero times the natural logarithm of two is equal to zero and π
raised to the zeroth power is also equal to one. So this simplifies to give us
negative 30 divided by the natural logarithm of two. Therefore, by using LβHΓ΄pitalβs
rule, weβve shown that the limit as π₯ approaches zero of negative six times the sin
of five π₯ all divided by two to the power of π₯ minus one is equal to negative 30
divided by the natural logarithm of two.