# Video: Comparing the Growth Rate of Two Functions Using Limits

Compare the growth rate of the two functions 𝑓(𝑥) = (𝑥 + 4)² and 𝑔(𝑥) = ln 𝑥 using limits as 𝑥 → ∞.

02:34

### Video Transcript

Compare the growth rate of the two functions 𝑓 of 𝑥 equals 𝑥 plus four squared and 𝑔 of 𝑥 equals the natural log of 𝑥 using limits as 𝑥 approaches ∞.

We begin by recalling that we can use limits to compare the growth rate of two functions by using the following definition. We ensure that 𝑓 of 𝑥 and 𝑔 of 𝑥 are positive for 𝑥 sufficiently large. Then if 𝑓 of 𝑥 grows faster than 𝑔 of 𝑥, the limit as 𝑥 approaches positive ∞ of the quotient 𝑓 of 𝑥 over 𝑔 of 𝑥 must itself be positive ∞. Or, equivalently, the limit as 𝑥 approaches positive ∞ of 𝑔 of 𝑥 over 𝑓 of 𝑥 would be equal to zero. Both 𝑓 of 𝑥 and 𝑔 of 𝑥 are positive for sufficiently large values of 𝑥. In fact, the natural log of 𝑥 is always positive and 𝑓 of 𝑥 is greater than or equal to zero for all values of 𝑥.

So we use the first part of our definition. We’ll find the limit as 𝑥 approaches positive ∞ of 𝑥 plus four all squared over the natural log of 𝑥. Now, we notice that if we try direct substitution, we obtain ∞ over ∞, which is, of course, of indeterminate form. Instead then, we’re going to use L’Hôpital’s rule. Now, this says that if 𝑎 is either a finite number or ∞, if the limit as 𝑥 approaches 𝑎 of the quotient of 𝑓 of 𝑥 and 𝑔 of 𝑥 is either equal to zero over zero or ∞ over ∞, then the limit as 𝑥 approaches 𝑎 of 𝑓 of 𝑥 over 𝑔 of 𝑥 is equal to the limit as 𝑥 approaches 𝑎 of 𝑓 prime of 𝑥 over 𝑔 prime of 𝑥.

So let’s begin by finding the derivative of 𝑓 and 𝑔. That’s 𝑓 prime of 𝑥 and 𝑔 prime of 𝑥. Now, the general power rule says that we can find the derivative of 𝑥 plus four all squared by multiplying 𝑥 plus four by two. And then reducing the exponent by one. And then multiplying all of that by the derivative of the inner function. Well, the derivative of 𝑥 plus four is one. So we have two times 𝑥 plus four to the power of one times one, which is simply two times 𝑥 plus four. the derivative of 𝑔, the derivative of the natural log of 𝑥, is a bit more straightforward. It’s simply one over 𝑥. And so, the limit as 𝑥 approaches positive ∞ of 𝑥 plus four all squared over the natural log of 𝑥 must be equal to the limit as 𝑥 approaches positive ∞ of two times 𝑥 plus four over one over 𝑥.

Now, of course, dividing by a fraction is the same as multiplying by the reciprocal of that fraction. So we get two times 𝑥 plus four times 𝑥 over one, which is just two 𝑥 times 𝑥 plus four. And so, we see that as 𝑥 approaches ∞, two 𝑥 times 𝑥 plus four also approaches ∞. And referring back to our original definition, we see that this means that 𝑓 of 𝑥 grows faster than 𝑔 of 𝑥. And so, the growth rate of 𝑓 of 𝑥 must be greater than the growth rate of 𝑔 of 𝑥.