### Video Transcript

For the functions π of π₯ equals
the natural log of two π₯ and π of π₯ equals four π₯ cubed, evaluate the limit as
π₯ approaches β of π of π₯ over π of π₯ using lβHΓ΄pitalβs rule.

Now, why might we want to do
this? Well, we can use limits to compare
the growth rate of two functions. We know that if π of π₯ and π of
π₯ are positive for sufficiently large values of π₯, then if the limit as π₯
approaches β of π of π₯ over π of π₯ is equal to β, π of π₯ grows faster than π
of π₯. Now, in this case, weβre actually
not looking to find the growth rate. But we are told to use lβHΓ΄pitalβs
rule to evaluate the limit.

Now, this says that if π is either
a finite number or β and if the limit as π₯ approaches π of π of π₯ over π of π₯
is either equal to zero over zero or the other indeterminate form β over β, then the
limit as π₯ approaches π of π of π₯ over π of π₯ can be calculated by working out
the limit as π₯ approaches π of π prime of π₯ over π prime of π₯. Now, in this question, weβre asked
to find the limit as π₯ approaches β of π of π₯ over π of π₯. Thatβs the natural log of two π₯
over four π₯ cubed. Well, this would give us the
indeterminate form β over β.

So to calculate the limit, what
weβre going to do is differentiate π and π with respect to π₯ and find the
quotient and then evaluate that limit as π₯ approaches β. Weβll start with π of π₯ equals
the natural log of two π₯. Now, we can quote a general result
for the derivative of the natural log of some constant times π₯. But letβs use a substitution to see
where it comes from. Weβre going to let π’ be equal to
two π₯. And of course, we know that the
derivative of π’ with respect to π₯ then will be equal to two.

The chain rule says that dπ¦ by dπ₯
must be equal to dπ¦ by dπ’ times dπ’ by dπ₯. So if we say that π¦ is equal to
the natural log of π’, we know that the derivative of π¦ with respect to π’ is one
over π’. So π prime of π₯ would be one over
π’ times two. But of course, we want this to be
in terms of π₯. So we replace π’ with our original
substitution with two π₯. And we can cancel this two. And we find that π prime of π₯ is
equal to one over π₯. And so, we can actually quote the
result that the derivative of the natural log of some constant times π₯ is simply
one over π₯.

π of π₯, of course, was four π₯
cubed. And we know that in this case, we
multiply the entire term by the exponent and reduce the exponent by one. And that gives us the derivative of
π with respect to π₯. Itβs three times four π₯ squared,
which is simply 12π₯ squared. And so, we have the limit as π₯
approaches β then of one over π₯ over 12π₯ squared. We can think about that as the same
as one over π₯ times one over 12π₯ squared, which would give us one over 12π₯
cubed.

And whatβs the limit as π₯
approaches β then of one over 12π₯ cubed? Well, as π₯ gets larger, one over
12π₯ cubed gets smaller. And so, as π₯ approaches β, one
over 12π₯ cubed approaches zero. And of course, we saw that the
limit as π₯ approaches β of one over π₯ over 12π₯ squared was equal to the original
limit. Thatβs the limit as π₯ approaches β
of the natural log of two π₯ by the four π₯ cubed.

Now, in fact, we have finished
answering the question. But itβs useful to spot what this
actually tells us about the growth rate of the function. It tells us that π of π₯ has a
slower growth rate than π of π₯. The function the natural log of two
π₯ grows slower than the function of four π₯ cubed.