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

Given that 𝑓(π‘₯) = √(5π‘₯) and 𝑔(π‘₯) = log₃ 5π‘₯, use lim_(π‘₯ β†’ ∞) 𝑓(π‘₯)/g(π‘₯) to determine whether 𝑓(π‘₯) or 𝑔(π‘₯) is dominant.

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Video Transcript

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

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

Finally, if the limit as π‘₯ approaches ∞ of 𝑓 of π‘₯ divided by 𝑔 of π‘₯ is equal to some nonzero real constant π‘Ž, 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 ∞ 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 ∞, and our denominator is also increasing without bound as π‘₯ approaches ∞. So, if we try to evaluate our limit directly, we get the indeterminate form ∞ divided by ∞. 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 L’HΓ΄pital’s rule.

We recall the following version of L’HΓ΄pital’s rule. If we have two differentiable functions 𝑓 and 𝑔, where the limit as π‘₯ approaches ∞ of 𝑓 of π‘₯ and the limit as π‘₯ approaches ∞ of 𝑔 of π‘₯ are both equal to ∞. Then we know the limit as π‘₯ approaches ∞ of 𝑓 of π‘₯ divided by 𝑔 of π‘₯ is equal to the limit as π‘₯ approaches ∞ of 𝑓 prime of π‘₯ divided by 𝑔 prime of π‘₯ provided that this limit exists or it’s equal to positive or negative ∞.

In other words, under these conditions, instead of evaluating the limit as π‘₯ approaches ∞ of 𝑓 of π‘₯ divided by 𝑔 of π‘₯, we can instead evaluate the limit as π‘₯ approaches ∞ of 𝑓 prime of π‘₯ divided by 𝑔 prime of π‘₯. So, let’s first check that we can actually use L’HΓ΄pital’s rule in this case. We first need to show that both 𝑓 and 𝑔 are differentiable functions. And we can do this directly in this case. We’ll start by finding an expression for 𝑓 prime of π‘₯.

First, 𝑓 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 𝑓 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, we’ve shown that 𝑓 is differentiable. Let’s now do the same for 𝑔.

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

The last thing we need to show to use this version of L’HΓ΄pital’s rule is the limit as π‘₯ approaches ∞ of 𝑓 of π‘₯ and the limit as π‘₯ approaches ∞ of 𝑔 of π‘₯ are both equal to ∞. And we’ve actually already shown both of these conditions are true when we originally tried to directly evaluate our limit. As π‘₯ approached ∞, both our numerator and our denominator grew without bound. In other words, the limit as π‘₯ approached ∞ of both 𝑓 of π‘₯ and 𝑔 of π‘₯ was equal to ∞.

So, we’ve shown all the conditions for this version of L’HΓ΄pital’s rule are true. So, to evaluate our limit, we can now instead attempt to evaluate the limit as π‘₯ approaches ∞ of 𝑓 prime of π‘₯ divided by 𝑔 prime of π‘₯. And we already found expressions for 𝑓 prime of π‘₯ and 𝑔 prime of π‘₯. So let’s write these in. This gives us the limit as π‘₯ approaches ∞ 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, let’s multiply it by the reciprocal. Doing this, we get the limit as π‘₯ approaches ∞ 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 ∞ 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 ∞, our numerator is increasing without bound. This means our limit is equal to ∞. And remember, we said, if this limit is equal to ∞, then we say that 𝑓 of π‘₯ is dominant.

Therefore, by using L’HΓ΄pital’s rule, we’ve shown if 𝑓 of π‘₯ is equal to the square root of five π‘₯ and 𝑔 of π‘₯ is equal to the log base three of five π‘₯. Then the limit as π‘₯ approaches ∞ of 𝑓 of π‘₯ divided by 𝑔 of π‘₯ is equal to ∞. And so, 𝑓 of π‘₯ is the dominant function.

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