Question Video: Determine Whether the Logarithm Function or Square Root Function Is Dominant | Nagwa Question Video: Determine Whether the Logarithm Function or Square Root Function Is Dominant | Nagwa

Question Video: Determine Whether the Logarithm Function or Square Root Function Is Dominant Mathematics • Higher Education

Given that π(π₯) = β(5π₯) and π(π₯) = logβ 5π₯, use lim_(π₯ β β) π(π₯)/g(π₯) to determine whether π(π₯) or π(π₯) is dominant.

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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₯. Letβs 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 LβHΓ΄pitalβs rule.

We recall the following version of LβHΓ΄pitalβs 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 itβs 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, letβs first check that we can actually use LβHΓ΄pitalβs 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. Weβll 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, weβve shown that π is differentiable. Letβs 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 ππ₯ 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 LβHΓ΄pitalβs 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 weβve 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, weβve shown all the conditions for this version of LβHΓ΄pitalβs 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 letβs 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, letβs 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 LβHΓ΄pitalβs rule, weβve 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.

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