Video Transcript
For the functions 𝑓 of 𝑥 equals
the natural log of two 𝑥 and 𝑔 of 𝑥 equals four 𝑥 cubed, evaluate the limit as
𝑥 approaches ∞ of 𝑓 of 𝑥 over 𝑔 of 𝑥 using l’Hôpital’s 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 𝑓 of 𝑥 and 𝑔 of
𝑥 are positive for sufficiently large values of 𝑥, then if the limit as 𝑥
approaches ∞ of 𝑓 of 𝑥 over 𝑔 of 𝑥 is equal to ∞, 𝑓 of 𝑥 grows faster than 𝑔
of 𝑥. Now, in this case, we’re actually
not looking to find the growth rate. But we are told to use l’Hôpital’s
rule to evaluate the limit.
Now, this says that if 𝑎 is either
a finite number or ∞ and if the limit as 𝑥 approaches 𝑎 of 𝑓 of 𝑥 over 𝑔 of 𝑥
is either equal to zero over zero or the other indeterminate form ∞ over ∞, then the
limit as 𝑥 approaches 𝑎 of 𝑓 of 𝑥 over 𝑔 of 𝑥 can be calculated by working out
the limit as 𝑥 approaches 𝑎 of 𝑓 prime of 𝑥 over 𝑔 prime of 𝑥. Now, in this question, we’re asked
to find the limit as 𝑥 approaches ∞ of 𝑓 of 𝑥 over 𝑔 of 𝑥. That’s the natural log of two 𝑥
over four 𝑥 cubed. Well, this would give us the
indeterminate form ∞ over ∞.
So to calculate the limit, what
we’re going to do is differentiate 𝑓 and 𝑔 with respect to 𝑥 and find the
quotient and then evaluate that limit as 𝑥 approaches ∞. We’ll start with 𝑓 of 𝑥 equals
the natural log of two 𝑥. Now, we can quote a general result
for the derivative of the natural log of some constant times 𝑥. But let’s use a substitution to see
where it comes from. We’re going to let 𝑢 be equal to
two 𝑥. And of course, we know that the
derivative of 𝑢 with respect to 𝑥 then will be equal to two.
The chain rule says that d𝑦 by d𝑥
must be equal to d𝑦 by d𝑢 times d𝑢 by d𝑥. So if we say that 𝑦 is equal to
the natural log of 𝑢, we know that the derivative of 𝑦 with respect to 𝑢 is one
over 𝑢. So 𝑓 prime of 𝑥 would be one over
𝑢 times two. But of course, we want this to be
in terms of 𝑥. So we replace 𝑢 with our original
substitution with two 𝑥. And we can cancel this two. And we find that 𝑓 prime of 𝑥 is
equal to one over 𝑥. And so, we can actually quote the
result that the derivative of the natural log of some constant times 𝑥 is simply
one over 𝑥.
𝑔 of 𝑥, of course, was four 𝑥
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
𝑔 with respect to 𝑥. It’s three times four 𝑥 squared,
which is simply 12𝑥 squared. And so, we have the limit as 𝑥
approaches ∞ then of one over 𝑥 over 12𝑥 squared. We can think about that as the same
as one over 𝑥 times one over 12𝑥 squared, which would give us one over 12𝑥
cubed.
And what’s the limit as 𝑥
approaches ∞ then of one over 12𝑥 cubed? Well, as 𝑥 gets larger, one over
12𝑥 cubed gets smaller. And so, as 𝑥 approaches ∞, one
over 12𝑥 cubed approaches zero. And of course, we saw that the
limit as 𝑥 approaches ∞ of one over 𝑥 over 12𝑥 squared was equal to the original
limit. That’s the limit as 𝑥 approaches ∞
of the natural log of two 𝑥 by the four 𝑥 cubed.
Now, in fact, we have finished
answering the question. But it’s useful to spot what this
actually tells us about the growth rate of the function. It tells us that 𝑓 of 𝑥 has a
slower growth rate than 𝑔 of 𝑥. The function the natural log of two
𝑥 grows slower than the function of four 𝑥 cubed.
