### 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. Weβll 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, weβre going to use
LβHΓ΄pitalβs 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 letβs begin by finding the
derivative of π and π. Thatβs 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. Itβs 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₯.