Video: L’Hopital’s Rule

In this video, we will learn how to apply L’Hopital’s rule to evaluate the limits of the indeterminate forms 0/0 and ∞/∞.

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Video Transcript

L’Hopital’s Rule.

In this video, we will learn how to apply L’Hopital’s rule to evaluate the limits of indeterminate forms zero over zero, infinity over infinity, and negative infinity over negative infinity. We’ll be looking at some examples of how we can use L’Hopital’s rule.

Let’s start by considering a limit. And that is the limit as 𝑥 tends to zero of 𝑥 over sin of five 𝑥. If we want to evaluate this limit, we can start by trying to use direct substitution. We obtain zero over sin of five times zero. Since five times zero is just zero, this is the same as zero over sin of zero. Sin of zero gives zero. So this must be equal to zero over zero, which is undefined. Therefore, this limit cannot be evaluated directly by using direct substitution.

In fact, any of the techniques we know so far cannot be used in order to evaluate this limit. This is where L’Hopital’s rule comes in. L’Hopital’s rule tells us that if the limit as 𝑥 approaches 𝑎 of some 𝑓 of 𝑥 over 𝑔 of 𝑥 is equal to zero over zero or the limit as 𝑥 approaches 𝑎 of 𝑓 of 𝑥 over 𝑔 of 𝑥 is equal to positive or negative infinity over positive or negative infinity. Where 𝑎 can be any real number, positive infinity, or negative infinity. Then the limit as 𝑥 approaches 𝑎 of 𝑓 of 𝑥 over 𝑔 of 𝑥 is equal to the limit as 𝑥 approaches 𝑎 of 𝑓 prime of 𝑥 over 𝑔 prime of 𝑥.

One thing to note with L’Hopital’s rule is that, in order for it to work, both 𝑓 of 𝑥 and 𝑔 of 𝑥 must be differentiable. And also with our condition, when we say that the limit as 𝑥 approaches 𝑎 of 𝑓 of 𝑥 over 𝑔 of 𝑥 can be equal to positive or negative infinity over positive or negative infinity. It must be equal to positive infinity over positive infinity or negative infinity over negative infinity. The rule does not work when the limit is equal to positive infinity over negative infinity or negative infinity over positive infinity.

Now that we’ve covered the definition of L’Hopital’s rule, let’s apply it to the limit we are trying to find. So that’s the limit as 𝑥 approaches zero of 𝑥 over sin of five 𝑥. And we’ve seen that using direct substitution, our limit is equal to zero over zero. Therefore, it satisfies the first condition of L’Hopital’s rule. We can see that the value of 𝑎 in our limit is zero. Since 𝑎 is a real number, it also satisfies the second condition. Now we can say that 𝑓 of 𝑥 is equal to 𝑥 and 𝑔 of 𝑥 is equal to sin of five 𝑥. Clearly, both of these functions, 𝑓 and 𝑔, are differentiable. Therefore, we’re ready to apply L’Hopital’s rule.

We first need to find 𝑓 prime of 𝑥 and 𝑔 prime of 𝑥. Differentiating 𝑥 with respect to 𝑥, we find that 𝑓 prime of 𝑥 is equal to one. In order to find 𝑔 prime of 𝑥, we need to differentiate sin of five 𝑥. This is a compound function. Therefore, we must use the chain rule. We differentiate the inside of the function — so that’s five 𝑥 — to get a constant of five. Then we differentiate the sine to get cos of five 𝑥, giving us that 𝑔 prime of 𝑥 is equal to five cos of five 𝑥.

Applying L’Hopital’s rule, we can say that the limit as 𝑥 approaches zero of 𝑥 over sin of five 𝑥 is equal to the limit as 𝑥 approaches zero of one over five cos of five 𝑥. And now we can simply use direct substitution. And we see that our limit is equal to one over five times cos of five times zero. Five times zero is simply zero. And cos of zero is simply one. Therefore, we can say that our limit is equal to one over five times one, which is simply one-fifth.

L’Hopital’s rule can be very useful for finding the limits of functions which seem as though they may not exist. Let’s look at an example.

Find the limit as 𝑥 approaches zero of seven 𝑒 to the five 𝑥 minus seven over negative 𝑒 to the eight 𝑥 plus one.

We will start by trying to solve this limit using direct substitution. We obtain seven times 𝑒 to the power of five times zero minus seven over negative 𝑒 to the power of eight times zero plus one. Since 𝑒 to the power of zero is equal to one, we find that this is equal to seven minus seven over negative one plus one, which simplifies to zero over zero. However, this is undefined. Although we obtain that our limit is undefined using direct substitution, it is equal to zero over zero. And this tells us that we may be able to use L’Hopital’s rule.

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

Now our limit satisfies the condition of our limit being equal to zero over zero. And since we’re taking the limit as 𝑥 approaches zero, that means that our 𝑎 is equal to zero, which is a real number. Therefore, we can use L’Hopital’s rule. 𝑓 of 𝑥 is the numerator of the function which we’re taking the limit of. So that’s seven 𝑒 to the five 𝑥 minus seven. And 𝑔 of 𝑥 is the denominator. So that’s negative 𝑒 to the power of eight 𝑥 plus one.

Now we must find 𝑓 prime of 𝑥 and 𝑔 prime of 𝑥. Since we will be differentiating exponential terms, we can use the rule that tells us that the differential of 𝑒 to the power of 𝑘𝑥 with respect to 𝑥 is equal to 𝑘 times 𝑒 to the power of 𝑘𝑥. Let’s differentiate 𝑓 of 𝑥 term by term. Seven 𝑒 to the power of five 𝑥 is an exponential term. So we’ll be using the rule which we’ve just stated. Our value of 𝑘 is five. And we notice that we have a constant of seven multiplying our exponential term. So that must remain two, giving us seven timesed by five 𝑒 to the five 𝑥. Then seven times five is 35. So we can write this as 35𝑒 to the power of five 𝑥.

The second term in 𝑓 of 𝑥 is negative seven, which is simply a constant. And when we differentiate any constant, we simply get zero. So we found that 𝑓 prime of 𝑥 is equal to 35𝑒 to the power of five 𝑥. The first term in 𝑔 of 𝑥 is negative 𝑒 to the power of eight 𝑥, which is again an exponential term. Using our rule, we obtain that the differential of this term is negative eight 𝑒 to the power of eight 𝑥. The second term in 𝑔 of 𝑥 is one, which is again a constant. And so this will differentiate to give zero.

We are now ready to apply L’Hopital’s rule. We obtain that the limit as 𝑥 approaches zero of seven 𝑒 to the power of five 𝑥 minus seven over negative 𝑒 to the power of eight 𝑥 plus one. Is equal to the limit as 𝑥 approaches zero of 35 times 𝑒 to the power of five 𝑥 over negative eight times 𝑒 to the power of eight 𝑥. And we can now apply direct substitution, giving us 35 timesed by 𝑒 to the power of zero over negative eight timesed by 𝑒 to the power of zero. Since 𝑒 to the power of zero is equal to one, we get a solution that our limit must be equal to negative 35 over eight.

Next, let’s look at an example which satisfies a different condition of L’Hopital’s rule.

Find the limit as 𝑥 tends to infinity of two timesed by 𝑒 to the power of three 𝑥 minus five over three times 𝑒 to the power of three 𝑥 minus one.

We can start by trying to find this limit using direct substitution. We’ll be using the fact that the limit as 𝑥 tends to infinity of 𝑒 to the power of 𝑥 is equal to positive infinity. From this, we obtain that the limit as 𝑥 tends to infinity of 𝑒 to the power of three 𝑥 is also equal to positive infinity. And this tells us that when we use direct substitution in order to find our limit, we find that it’s equal to positive infinity over positive infinity. And this is undefined. Therefore, we’ve not yet found a solution.

However, the fact that it is equal to positive infinity over positive infinity does tell us that we’re able to use L’Hopital’s rule. L’Hopital’s rule tells us that if the limit as 𝑥 approaches 𝑎 of 𝑓 of 𝑥 over 𝑔 of 𝑥 is equal to zero over zero, positive infinity over positive infinity, or negative infinity over negative infinity. Where 𝑎 is a real number, positive infinity, or negative infinity. Then the limit as 𝑥 approaches 𝑎 of 𝑓 of 𝑥 over 𝑔 of 𝑥 is equal to the limit as 𝑥 approaches 𝑎 of 𝑓 prime of 𝑥 over 𝑔 prime of 𝑥.

Now our limit is equal to infinity over infinity. And we’re taking the limit as 𝑥 tends to positive infinity. Therefore, we’re allowed to use L’Hopital’s rule. In our case, 𝑓 of 𝑥 is equal to two times 𝑒 to the power of three 𝑥 minus five. And 𝑔 of 𝑥 is equal to three timesed by 𝑒 to the power of three 𝑥 minus one.

We find 𝑓 prime and 𝑔 prime by differentiating 𝑓 and 𝑔. Differentiating two 𝑒 to the power of three 𝑥 minus five with respect to 𝑥, we obtain that 𝑓 prime of 𝑥 must be equal to six 𝑒 to the power of three 𝑥. And differentiating three times 𝑒 to the power of three 𝑥 minus one with respect to 𝑥, we obtain that 𝑔 prime of 𝑥 must be equal to nine 𝑒 to the three 𝑥. We obtain that our limit must be equal to the limit as 𝑥 tends to infinity of six timesed by 𝑒 to the three 𝑥 over nine timesed by 𝑒 to the three 𝑥.

Here we notice that we have a factor of three timesed by 𝑒 to the three 𝑥 in both the numerator and denominator. Since we can write our numerator as two timesed by three 𝑒 to the power of three 𝑥 and our denominator as three timesed by three 𝑒 to the power of three 𝑥. Therefore, these factors of three timesed by 𝑒 to the power of three 𝑥 will cancel out, leaving us with the limit as 𝑥 tends to infinity of two over three. Since there’s no 𝑥 dependency within our limit, our limit is simply equal to two-thirds. And this is the solution to the question.

Many different limits give one of the indeterminate forms required to use L’Hopital’s rule. Let’s consider the following examples.

Determine the limit as 𝑥 approaches one of negative 11 times the natural logarithm of 𝑥 over negative nine 𝑥 plus nine.

Let’s start by trying to find this limit using direct substitution. We obtain negative 11 multiplied by the natural logarithm of one over negative nine plus nine. We will use the fact that the natural logarithm of one is equal to zero. And this gives us that our limit must be equal to zero over zero, which is undefined. However, this is the main condition which we must satisfy in order to use L’Hopital’s rule.

This is L’Hopital’s rule. And since our limit is equal to zero over zero, we can see that we have satisfied the first condition. We’re taking the limit as 𝑥 approaches one. So we can say that 𝑎 is equal to one. And one is a real number. Therefore, we have satisfied the second condition. And this tells us that we’re able to use L’Hopital’s rule. We can say that 𝑓 of 𝑥 is equal to negative 11 timesed by the natural logarithm of 𝑥. And 𝑔 of 𝑥 is equal to negative nine 𝑥 plus nine.

In order to differentiate 𝑓 with respect to 𝑥, we’ll be using the fact that the differential of the natural logarithm of 𝑥 with respect to 𝑥 is equal to one over 𝑥. Since 𝑓 of 𝑥 is simply a constant multiplied by the natural logarithm of 𝑥, where the constant is equal to negative 11, we obtain that 𝑓 prime of 𝑥 is equal to negative 11 over 𝑥. Now 𝑔 of 𝑥 is equal to negative nine 𝑥 plus nine, which is simply a polynomial. And so we can differentiate this using the power rule for differentiation, giving us that 𝑔 prime of 𝑥 is equal to negative nine.

Now we can apply the formula for L’Hopital’s rule, which is that the limit as 𝑥 approaches 𝑎 of 𝑓 of 𝑥 over 𝑔 of 𝑥 is equal to the limit as 𝑥 approaches 𝑎 of 𝑓 prime of 𝑥 over 𝑔 prime of 𝑥. Which tells us that the limit as 𝑥 approaches one of negative 11 times the natural logarithm of 𝑥 over negative nine plus nine. Is equal to the limit as 𝑥 approaches one of negative 11 over negative nine 𝑥. We see that we have a factor of negative one in both the numerator and denominator. And so they can be cancelled, giving us the limit as 𝑥 approaches one of 11 over nine 𝑥. And here we can apply direct substitution, giving us 11 over nine times one. This gives us the solution that our limit must be equal to 11 over nine.

In the next example, we’ll be seeing how we can use L’Hopital’s rule in order to find another result.

Given functions lowercase 𝑓 and capital 𝐹 that are positive for large values of 𝑥, we say that capital 𝐹 dominates lowercase 𝑓 as 𝑥 tends to infinity if the limit as 𝑥 tends to infinity of lowercase 𝑓 of 𝑥 over capital 𝐹 of 𝑥 is equal to zero. Use L’Hopital’s rule to determine which is dominant as 𝑥 tends to infinity. The natural logarithm of 𝑥 or the square root of 𝑥.

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

Let’s start by considering the latter of these two options. We need to find the limit as 𝑥 tends to infinity of the square root of 𝑥 over the natural logarithm of 𝑥. Since the square root of 𝑥 and the natural logarithm of 𝑥 are both increasing functions, we know that the limit of each of these individually as 𝑥 tends to positive infinity will be positive infinity. Therefore, the limit as 𝑥 tends to infinity of the square root of 𝑥 over the natural logarithm of 𝑥 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 𝑥 approaches 𝑎 of 𝑓 of 𝑥 over 𝑔 of 𝑥 is equal to zero over zero, positive infinity over positive infinity, or negative infinity over negative infinity. Where 𝑎 is a real number, positive infinity, or negative infinity. Then the limit as 𝑥 approaches 𝑎 of 𝑓 of 𝑥 over 𝑔 of 𝑥 is equal to the limit as 𝑥 approaches 𝑎 of 𝑓 prime of 𝑥 over 𝑔 prime of 𝑥.

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 𝑥 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 𝑥 is equal to the square root of 𝑥 and 𝑔 of 𝑥 is equal to the natural logarithm of 𝑥. The square root of 𝑥 is also equal to 𝑥 to the power of a half. And so in order to find 𝑓 prime of 𝑥, 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 𝑥 is equal to one-half multiplied by 𝑥 to the power of negative one-half.

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

Simplifying this, we obtain the limit as 𝑥 tends to infinity of 𝑥 over two timesed by 𝑥 to the power of one-half. Now we can cancel a factor of 𝑥 to the power of one-half from the top and bottom. We obtain the limit as 𝑥 tends to infinity of 𝑥 to the power of one-half over two. And the only 𝑥 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 𝑥 does not dominate the square root of 𝑥.

Now let’s check whether the limit as 𝑥 tends to infinity of the natural logarithm of 𝑥 over the square root of 𝑥 is equal to zero. Now the natural logarithm of 𝑥 over the square root of 𝑥 is the reciprocal of the square root of 𝑥 over the natural logarithm of 𝑥. And so when we use direct substitution in order to find the limit as 𝑥 tends to infinity, we will again get infinity over infinity. So we can say that the limit as 𝑥 tends to infinity of the natural logarithm of 𝑥 over the square root of 𝑥 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 𝑥 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 𝑥 tends to infinity of the square root of 𝑥 over the natural logarithm of 𝑥 is equal to the limit as 𝑥 tends to infinity of 𝑥 to the power of half over two. This tells us that the limit as 𝑥 tends to infinity of the natural logarithm of 𝑥 over the square root of 𝑥 will be equal to the limit as 𝑥 tends to infinity of the reciprocal of 𝑥 to the power of one-half over two. And that is the limit as 𝑥 tends to infinity of two over 𝑥 to the power of one-half.

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

As we have seen from this example, L’Hopital’s rule can be useful for showing almost anything involving a limit, since it enables us to find limits which would otherwise be undefined. We’ve now seen a variety of examples involving L’Hopital’s rule. Let’s look at some key points of the video. Key Points. L’Hopital’s rule can be used to find limits of the indeterminate forms zero over zero, positive infinity over positive infinity, and negative infinity over negative infinity. L’Hopital’s rule is as follows. Suppose the limit as 𝑥 approaches 𝑎 of 𝑓 of 𝑥 over 𝑔 of 𝑥 equals zero over zero. Or the limit as 𝑥 approaches 𝑎 of 𝑓 of 𝑥 over 𝑔 of 𝑥 is equal to positive or negative infinity over positive or negative infinity. Where 𝑎 can be any real number, positive infinity, or negative infinity. Then the limit as 𝑥 approaches 𝑎 of 𝑓 of 𝑥 over 𝑔 of 𝑥 is equal to the limit as 𝑥 approaches 𝑎 of 𝑓 prime of 𝑥 over 𝑔 prime of 𝑥.

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