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# Question Video: Comparing the Growth Rate of Two Functions Using Limits Mathematics • Higher Education

Compare the growth rate of the two functions π(π₯) = (π₯ + 4)Β² and π(π₯) = ln π₯ using limits as π₯ β β.

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### Video Transcript

Compare the growth rate of the two functions π of π₯ equals π₯ plus four squared and π of π₯ equals the natural log of π₯ using limits as π₯ approaches β.

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 π of π₯ and π of π₯ are positive for π₯ sufficiently large. Then if π of π₯ grows faster than π of π₯, the limit as π₯ approaches positive β of the quotient π of π₯ over π of π₯ must itself be positive β. Or, equivalently, the limit as π₯ approaches positive β of π of π₯ over π of π₯ would be equal to zero. Both π of π₯ and π of π₯ are positive for sufficiently large values of π₯. In fact, the natural log of π₯ is always positive and π of π₯ is greater than or equal to zero for all values of π₯.

So we use the first part of our definition. Weβll find the limit as π₯ approaches positive β of π₯ plus four all squared over the natural log of π₯. Now, we notice that if we try direct substitution, we obtain β over β, which is, of course, of indeterminate form. Instead then, weβre going to use LβHΓ΄pitalβs rule. Now, this says that if π is either a finite number or β, if the limit as π₯ approaches π of the quotient of π of π₯ and π of π₯ is either equal to zero over zero or β over β, then the limit as π₯ approaches π of π of π₯ over π of π₯ is equal to the limit as π₯ approaches π of π prime of π₯ over π prime of π₯.

So letβs begin by finding the derivative of π and π. Thatβs π prime of π₯ and π 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 π, the derivative of the natural log of π₯, is a bit more straightforward. Itβs simply one over π₯. And so, the limit as π₯ approaches positive β of π₯ plus four all squared over the natural log of π₯ must be equal to the limit as π₯ approaches positive β 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 β, two π₯ times π₯ plus four also approaches β. And referring back to our original definition, we see that this means that π of π₯ grows faster than π of π₯. And so, the growth rate of π of π₯ must be greater than the growth rate of π of π₯.

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