# Video: Find the Limit of a Quotient Using L’Hopital’s Rule

For the functions 𝑓(𝑥) = ln (2𝑥) and 𝑔(𝑥) = 4𝑥³, evaluate lim_(𝑥 →∞) (𝑓(𝑥)/𝑔(𝑥)) using l’Hôpital’s rule.

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### 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.