Video: Comparing the Growth Rate of Two Functions Using Limits

Compare the growth rate of the two functions 𝑓(π‘₯) = (π‘₯ + 4)Β² and 𝑔(π‘₯) = ln π‘₯ using limits as π‘₯ β†’ ∞.

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Video Transcript

Compare the growth rate of the two functions 𝑓 of π‘₯ equals π‘₯ plus four squared and 𝑔 of π‘₯ equals the natural log of π‘₯ using limits as π‘₯ 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 π‘₯ and 𝑔 of π‘₯ are positive for π‘₯ sufficiently large. Then if 𝑓 of π‘₯ grows faster than 𝑔 of π‘₯, the limit as π‘₯ approaches positive ∞ of the quotient 𝑓 of π‘₯ over 𝑔 of π‘₯ must itself be positive ∞. Or, equivalently, the limit as π‘₯ approaches positive ∞ of 𝑔 of π‘₯ over 𝑓 of π‘₯ would be equal to zero. Both 𝑓 of π‘₯ and 𝑔 of π‘₯ are positive for sufficiently large values of π‘₯. In fact, the natural log of π‘₯ is always positive and 𝑓 of π‘₯ is greater than or equal to zero for all values of π‘₯.

So we use the first part of our definition. We’ll find the limit as π‘₯ approaches positive ∞ of π‘₯ plus four all squared over the natural log of π‘₯. 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’HΓ΄pital’s rule. Now, this says that if π‘Ž is either a finite number or ∞, if the limit as π‘₯ approaches π‘Ž of the quotient of 𝑓 of π‘₯ and 𝑔 of π‘₯ is either equal to zero over zero or ∞ over ∞, then the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ over 𝑔 of π‘₯ is equal to the limit as π‘₯ approaches π‘Ž of 𝑓 prime of π‘₯ over 𝑔 prime of π‘₯.

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

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

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