### Video Transcript

Given functions lowercase π and
capital πΉ that are positive for large values of π₯, we say that capital πΉ
dominates lowercase π as π₯ tends to infinity if the limit as π₯ tends to infinity
of lowercase π of π₯ over capital πΉ of π₯ is equal to zero. Use LβHopitalβs rule to determine
which is dominant as π₯ tends to infinity. The natural logarithm of π₯ or the
square root of π₯.

Using the definition of dominant
given in the question, in order to say whether the natural logarithm of π₯ or the
square root of π₯ is dominant. We need to show that either the
limit as π₯ tends to infinity of the natural logarithm of π₯ over the square root of
π₯ is equal to zero. Or the limit as π₯ tends to
infinity of the square root of π₯ over the natural logarithm of π₯ is equal to
zero.

Letβs start by considering the
latter of these two options. We need to find the limit as π₯
tends to infinity of the square root of π₯ over the natural logarithm of π₯. Since the square root of π₯ and the
natural logarithm of π₯ are both increasing functions, we know that the limit of
each of these individually as π₯ tends to positive infinity will be positive
infinity. Therefore, the limit as π₯ tends to
infinity of the square root of π₯ over the natural logarithm of π₯ is equal to
infinity over infinity. And this is undefined. However, it does give us the main
condition for using LβHopitalβs rule.

LβHopitalβs rule tells us that if
the limit as π₯ approaches π of π of π₯ over π of π₯ is equal to zero over zero,
positive infinity over positive infinity, or negative infinity over negative
infinity. Where π is a real number, positive
infinity, or negative infinity. 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₯.

Now since our limit is equal to
positive infinity over positive infinity, we have satisfied the first condition. And weβre taking the limit as π₯
tends to positive infinity. Therefore, weβve also satisfied the
second condition. So weβre able to use LβHopitalβs
rule.

We have that π of π₯ is equal to
the square root of π₯ and π of π₯ is equal to the natural logarithm of π₯. The square root of π₯ is also equal
to π₯ to the power of a half. And so in order to find π prime of
π₯, weβll use the power rule for differentiation. We multiply by the power and
decrease the power by one, giving us that π prime of π₯ is equal to one-half
multiplied by π₯ to the power of negative one-half.

In order to differentiate π of π₯
with respect to π₯, we use the fact that the differential of the natural logarithm
of π₯ with respect to π₯ is equal to one over π₯. And so π prime of π₯ is equal to
one over π₯. Applying LβHopitalβs rule, we find
that our limit is equal to the limit as π₯ tends to infinity of π₯ to the power of
negative one-half over two timesed by one over π₯.

Simplifying this, we obtain the
limit as π₯ tends to infinity of π₯ over two timesed by π₯ to the power of
one-half. Now we can cancel a factor of π₯ to
the power of one-half from the top and bottom. We obtain the limit as π₯ tends to
infinity of π₯ to the power of one-half over two. And the only π₯ term here has a
positive power. And itβs in the numerator of the
fraction. Therefore, this limit must be equal
to infinity. And so it is not equal to zero. And we can conclude from this that
the natural logarithm of π₯ does not dominate the square root of π₯.

Now letβs check whether the limit as
π₯ tends to infinity of the natural logarithm of π₯ over the square root of π₯ is
equal to zero. Now the natural logarithm of π₯ over
the square root of π₯ is the reciprocal of the square root of π₯ over the natural
logarithm of π₯. And so when we use direct
substitution in order to find the limit as π₯ tends to infinity, we will again get
infinity over infinity. So we can say that the limit as π₯
tends to infinity of the natural logarithm of π₯ over the square root of π₯ must be
equal to infinity over infinity, which is again undefined. However, it allowed us to use
LβHopitalβs rule, since these two conditions are satisfied.

Our limit is equal to positive
infinity over positive infinity. And weβre taking the limit as π₯
tends to positive infinity. Since weβre taking the limit of the
reciprocal function, our π and π will be the other way around. Therefore, for our final line of
working, we will simply be taking the limit of the reciprocal function. And since the limit as π₯ tends to
infinity of the square root of π₯ over the natural logarithm of π₯ is equal to the
limit as π₯ tends to infinity of π₯ to the power of half over two. This tells us that the limit as π₯
tends to infinity of the natural logarithm of π₯ over the square root of π₯ will be
equal to the limit as π₯ tends to infinity of the reciprocal of π₯ to the power of
one-half over two. And that is the limit as π₯ tends to
infinity of two over π₯ to the power of one-half.

In this limit, our π₯ has a
positive power of one-half. However, itβs in the denominator of
the fraction. And therefore, if we take the limit
as π₯ tends to infinity, it will be equal to zero. And so we have shown that the limit
as π₯ tends to infinity of the natural logarithm of π₯ over the square root of π₯ is
equal to zero. From this, we can conclude that the
square root of π₯ dominates the natural logarithm of π₯.