### Video Transcript

Given that π of π₯ is equal to the square root of five π₯ and π of π₯ is equal to the log base three of five π₯, use the limit as π₯ approaches β of π of π₯ divided by π of π₯ to determine whether π of π₯ or π of π₯ is dominant.

The question gives us two functions, π of π₯ and π of π₯. The question wants us to determine which of these two functions is dominant by evaluating the limit as π₯ approaches β of π of π₯ divided by π of π₯. Letβs first recall what it means for a function to be dominant. We recall for eventually positive functions π and π, if the limit as π₯ approaches β of π of π₯ divided by π of π₯ is equal to β, then we say the function π of π₯ is dominant. However, if the limit as π₯ approaches β of π of π₯ divided by π of π₯ is equal to zero, then we say that π of π₯ is dominant.

Finally, if the limit as π₯ approaches β of π of π₯ divided by π of π₯ is equal to some nonzero real constant π, then we say that neither of these functions are dominant. So, to determine which of our functions is dominant, we need to evaluate the limit as π₯ approaches β of the square root of five π₯ divided by the log base three of five π₯.

However, if we try to evaluate this limit directly, we run into a problem. Our numerator is increasing without bound as π₯ approaches β, and our denominator is also increasing without bound as π₯ approaches β. So, if we try to evaluate our limit directly, we get the indeterminate form β divided by β. We recall if we evaluate the limit of a quotient of two functions and we get an indeterminant form, one way of attempting to deal with this is by using LβHΓ΄pitalβs rule.

We recall the following version of LβHΓ΄pitalβs rule. If we have two differentiable functions π and π, where the limit as π₯ approaches β of π of π₯ and the limit as π₯ approaches β of π of π₯ are both equal to β. Then we know the limit as π₯ approaches β of π of π₯ divided by π of π₯ is equal to the limit as π₯ approaches β of π prime of π₯ divided by π prime of π₯ provided that this limit exists or itβs equal to positive or negative β.

In other words, under these conditions, instead of evaluating the limit as π₯ approaches β of π of π₯ divided by π of π₯, we can instead evaluate the limit as π₯ approaches β of π prime of π₯ divided by π prime of π₯. So, letβs first check that we can actually use LβHΓ΄pitalβs rule in this case. We first need to show that both π and π are differentiable functions. And we can do this directly in this case. Weβll start by finding an expression for π prime of π₯.

First, π of π₯ is equal to the square root of five times π₯ to the power of one-half. Then, we can differentiate this by using the power rule for differentiation. We multiply by the exponent and reduce the exponent by one. This gives us π prime of π₯ is equal to one-half times root five times π₯ to the power of negative one-half. And by using our laws of exponents, we can rewrite this as root five divided by two root π₯. So, weβve shown that π is differentiable. Letβs now do the same for π.

We have that π of π₯ is equal to the log base three of five π₯. And to differentiate this, we recall for positive constants π and π, where π is not equal to one, the derivative of the log base π of ππ₯ with respect to π₯ is equal to one divided by π₯ times the natural logarithm of π. Using this, we get π prime of π₯ is equal to one divided by π₯ times the natural logarithm of three. So, π of π₯ is a differentiable function.

The last thing we need to show to use this version of LβHΓ΄pitalβs rule is the limit as π₯ approaches β of π of π₯ and the limit as π₯ approaches β of π of π₯ are both equal to β. And weβve actually already shown both of these conditions are true when we originally tried to directly evaluate our limit. As π₯ approached β, both our numerator and our denominator grew without bound. In other words, the limit as π₯ approached β of both π of π₯ and π of π₯ was equal to β.

So, weβve shown all the conditions for this version of LβHΓ΄pitalβs rule are true. So, to evaluate our limit, we can now instead attempt to evaluate the limit as π₯ approaches β of π prime of π₯ divided by π prime of π₯. And we already found expressions for π prime of π₯ and π prime of π₯. So letβs write these in. This gives us the limit as π₯ approaches β of the square root of five divided by two root π₯ all divided by one divided by π₯ times the natural logarithm of three.

To help us evaluate this limit, instead of dividing by the fraction one divided by π₯ times the natural logarithm of three, letβs multiply it by the reciprocal. Doing this, we get the limit as π₯ approaches β of root five times π₯ times the natural logarithm of three divided by two root π₯. And we now see both our numerator and our denominator share a factor of the square root of π₯. If we cancel this shared factor, we get the limit as π₯ approaches β of root five times root π₯ times the natural logarithm of three divided by two.

And now, we can evaluate this limit directly. The only part of this limit which changes as the value of π₯ changes is the square root of π₯. Everything else is a positive constant. So, as π₯ approaches β, our numerator is increasing without bound. This means our limit is equal to β. And remember, we said, if this limit is equal to β, then we say that π of π₯ is dominant.

Therefore, by using LβHΓ΄pitalβs rule, weβve shown if π of π₯ is equal to the square root of five π₯ and π of π₯ is equal to the log base three of five π₯. Then the limit as π₯ approaches β of π of π₯ divided by π of π₯ is equal to β. And so, π of π₯ is the dominant function.