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

For the functions 𝑓(π‘₯) = ln (2π‘₯) and 𝑔(π‘₯) = 4π‘₯Β³, evaluate lim_(π‘₯ β†’βˆž) (𝑓(π‘₯)/𝑔(π‘₯)) using l’HΓ΄pital’s rule.

03:11

Video Transcript

For the functions 𝑓 of π‘₯ equals the natural log of two π‘₯ and 𝑔 of π‘₯ equals four π‘₯ cubed, evaluate the limit as π‘₯ approaches ∞ of 𝑓 of π‘₯ over 𝑔 of π‘₯ using l’HΓ΄pital’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 π‘₯ and 𝑔 of π‘₯ are positive for sufficiently large values of π‘₯, then if the limit as π‘₯ approaches ∞ of 𝑓 of π‘₯ over 𝑔 of π‘₯ is equal to ∞, 𝑓 of π‘₯ grows faster than 𝑔 of π‘₯. Now, in this case, we’re actually not looking to find the growth rate. But we are told to use l’HΓ΄pital’s rule to evaluate the limit.

Now, this says that if π‘Ž is either a finite number or ∞ and if the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ over 𝑔 of π‘₯ is either equal to zero over zero or the other indeterminate form ∞ over ∞, then the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ over 𝑔 of π‘₯ can be calculated by working out the limit as π‘₯ approaches π‘Ž of 𝑓 prime of π‘₯ over 𝑔 prime of π‘₯. Now, in this question, we’re asked to find the limit as π‘₯ approaches ∞ of 𝑓 of π‘₯ over 𝑔 of π‘₯. That’s the natural log of two π‘₯ over four π‘₯ 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 π‘₯ and find the quotient and then evaluate that limit as π‘₯ approaches ∞. We’ll start with 𝑓 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 let’s use a substitution to see where it comes from. We’re going to let 𝑒 be equal to two π‘₯. And of course, we know that the derivative of 𝑒 with respect to π‘₯ then will be equal to two.

The chain rule says that d𝑦 by dπ‘₯ must be equal to d𝑦 by d𝑒 times d𝑒 by dπ‘₯. So if we say that 𝑦 is equal to the natural log of 𝑒, we know that the derivative of 𝑦 with respect to 𝑒 is one over 𝑒. So 𝑓 prime of π‘₯ would be one over 𝑒 times two. But of course, we want this to be in terms of π‘₯. So we replace 𝑒 with our original substitution with two π‘₯. And we can cancel this two. And we find that 𝑓 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‘₯.

𝑔 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 𝑔 with respect to π‘₯. It’s three times four π‘₯ squared, which is simply 12π‘₯ squared. And so, we have the limit as π‘₯ approaches ∞ 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 what’s the limit as π‘₯ approaches ∞ then of one over 12π‘₯ cubed? Well, as π‘₯ gets larger, one over 12π‘₯ cubed gets smaller. And so, as π‘₯ approaches ∞, one over 12π‘₯ cubed approaches zero. And of course, we saw that the limit as π‘₯ approaches ∞ of one over π‘₯ over 12π‘₯ squared was equal to the original limit. That’s the limit as π‘₯ approaches ∞ of the natural log of two π‘₯ by the four π‘₯ 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 π‘₯ has a slower growth rate than 𝑔 of π‘₯. The function the natural log of two π‘₯ grows slower than the function of four π‘₯ cubed.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.