Video: Using L’HΓ΄pital’s Rule to Find the Value of a Limit Involving a Logarithmic Function at Infinity

Given functions 𝑓 and 𝐹, that are positive for large values of π‘₯, we say that 𝐹 dominates 𝑓 as π‘₯ β†’ ∞ if lim_(π‘₯ β†’ ∞) 𝑓((π‘₯))/(𝐹(π‘₯)) = 0. Use L’Hopital’s rule to determine which is dominant as π‘₯ β†’ ∞: ln π‘₯ or √π‘₯.

04:48

Video Transcript

Given functions lowercase 𝑓 and capital 𝐹 that are positive for large values of π‘₯, we say that capital 𝐹 dominates lowercase 𝑓 as π‘₯ tends to infinity if the limit as π‘₯ tends to infinity of lowercase 𝑓 of π‘₯ over capital 𝐹 of π‘₯ is equal to zero. Use L’Hopital’s 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.

Let’s 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 L’Hopital’s rule.

L’Hopital’s rule tells us that if the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ over 𝑔 of π‘₯ is equal to zero over zero, positive infinity over positive infinity, or negative infinity over negative infinity. Where π‘Ž is a real number, positive infinity, or negative infinity. Then the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ over 𝑔 of π‘₯ is equal to the limit as π‘₯ approaches π‘Ž of 𝑓 prime of π‘₯ over 𝑔 prime of π‘₯.

Now since our limit is equal to positive infinity over positive infinity, we have satisfied the first condition. And we’re taking the limit as π‘₯ tends to positive infinity. Therefore, we’ve also satisfied the second condition. So we’re able to use L’Hopital’s rule.

We have that 𝑓 of π‘₯ is equal to the square root of π‘₯ and 𝑔 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 𝑓 prime of π‘₯, we’ll use the power rule for differentiation. We multiply by the power and decrease the power by one, giving us that 𝑓 prime of π‘₯ is equal to one-half multiplied by π‘₯ to the power of negative one-half.

In order to differentiate 𝑔 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 𝑔 prime of π‘₯ is equal to one over π‘₯. Applying L’Hopital’s 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 it’s 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 let’s 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 L’Hopital’s rule, since these two conditions are satisfied.

Our limit is equal to positive infinity over positive infinity. And we’re taking the limit as π‘₯ tends to positive infinity. Since we’re taking the limit of the reciprocal function, our 𝑓 and 𝑔 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, it’s 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‘₯.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.