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L’Hopital’s Rule.
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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.
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We’ll be looking at some examples of how we can use L’Hopital’s rule.
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Let’s start by considering a limit.
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And that is the limit as 𝑥 tends to zero of 𝑥 over sin of five 𝑥.
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If we want to evaluate this limit, we can start by trying to use direct substitution.
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We obtain zero over sin of five times zero.
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Since five times zero is just zero, this is the same as zero over sin of zero.
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Sin of zero gives zero.
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So this must be equal to zero over zero, which is undefined.
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Therefore, this limit cannot be evaluated directly by using direct substitution.
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In fact, any of the techniques we know so far cannot be used in order to evaluate this limit.
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This is where L’Hopital’s rule comes in.
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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.
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Where 𝑎 can be any real number, positive infinity, or negative infinity.
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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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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.
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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.
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It must be equal to positive infinity over positive infinity or negative infinity over negative infinity.
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The rule does not work when the limit is equal to positive infinity over negative infinity or negative infinity over positive infinity.
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Now that we’ve covered the definition of L’Hopital’s rule, let’s apply it to the limit we are trying to find.
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So that’s the limit as 𝑥 approaches zero of 𝑥 over sin of five 𝑥.
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And we’ve seen that using direct substitution, our limit is equal to zero over zero.
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Therefore, it satisfies the first condition of L’Hopital’s rule.
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We can see that the value of 𝑎 in our limit is zero.
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Since 𝑎 is a real number, it also satisfies the second condition.
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Now we can say that 𝑓 of 𝑥 is equal to 𝑥 and 𝑔 of 𝑥 is equal to sin of five 𝑥.
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Clearly, both of these functions, 𝑓 and 𝑔, are differentiable.
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Therefore, we’re ready to apply L’Hopital’s rule.
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We first need to find 𝑓 prime of 𝑥 and 𝑔 prime of 𝑥.
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Differentiating 𝑥 with respect to 𝑥, we find that 𝑓 prime of 𝑥 is equal to one.
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In order to find 𝑔 prime of 𝑥, we need to differentiate sin of five 𝑥.
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This is a compound function.
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Therefore, we must use the chain rule.
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We differentiate the inside of the function — so that’s five 𝑥 — to get a constant of five.
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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.
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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.
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And now we can simply use direct substitution.
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And we see that our limit is equal to one over five times cos of five times zero.
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Five times zero is simply zero.
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And cos of zero is simply one.
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Therefore, we can say that our limit is equal to one over five times one, which is simply one-fifth.
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L’Hopital’s rule can be very useful for finding the limits of functions which seem as though they may not exist.
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Let’s look at an example.
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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.
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We will start by trying to solve this limit using direct substitution.
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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.
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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.
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However, this is undefined.
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Although we obtain that our limit is undefined using direct substitution, it is equal to zero over zero.
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And this tells us that we may be able to use L’Hopital’s rule.
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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.
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Where 𝑎 is a real number, positive infinity, or negative infinity.
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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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Now our limit satisfies the condition of our limit being equal to zero over zero.
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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.
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Therefore, we can use L’Hopital’s rule.
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𝑓 of 𝑥 is the numerator of the function which we’re taking the limit of.
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So that’s seven 𝑒 to the five 𝑥 minus seven.
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And 𝑔 of 𝑥 is the denominator.
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So that’s negative 𝑒 to the power of eight 𝑥 plus one.
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Now we must find 𝑓 prime of 𝑥 and 𝑔 prime of 𝑥.
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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.
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Let’s differentiate 𝑓 of 𝑥 term by term.
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Seven 𝑒 to the power of five 𝑥 is an exponential term.
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So we’ll be using the rule which we’ve just stated.
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Our value of 𝑘 is five.
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And we notice that we have a constant of seven multiplying our exponential term.
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So that must remain two, giving us seven timesed by five 𝑒 to the five 𝑥.
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Then seven times five is 35.
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So we can write this as 35𝑒 to the power of five 𝑥.
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The second term in 𝑓 of 𝑥 is negative seven, which is simply a constant.
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And when we differentiate any constant, we simply get zero.
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So we found that 𝑓 prime of 𝑥 is equal to 35𝑒 to the power of five 𝑥.
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The first term in 𝑔 of 𝑥 is negative 𝑒 to the power of eight 𝑥, which is again an exponential term.
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Using our rule, we obtain that the differential of this term is negative eight 𝑒 to the power of eight 𝑥.
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The second term in 𝑔 of 𝑥 is one, which is again a constant.
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And so this will differentiate to give zero.
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We are now ready to apply L’Hopital’s rule.
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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.
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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.
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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.
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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.
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Next, let’s look at an example which satisfies a different condition of L’Hopital’s rule.
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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.
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We can start by trying to find this limit using direct substitution.
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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.
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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.
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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.
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And this is undefined.
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Therefore, we’ve not yet found a solution.
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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.
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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.
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Where 𝑎 is a real number, positive infinity, or negative infinity.
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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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Now our limit is equal to infinity over infinity.
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And we’re taking the limit as 𝑥 tends to positive infinity.
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Therefore, we’re allowed to use L’Hopital’s rule.
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In our case, 𝑓 of 𝑥 is equal to two times 𝑒 to the power of three 𝑥 minus five.
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And 𝑔 of 𝑥 is equal to three timesed by 𝑒 to the power of three 𝑥 minus one.
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We find 𝑓 prime and 𝑔 prime by differentiating 𝑓 and 𝑔.
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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.
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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.
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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.
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Here we notice that we have a factor of three timesed by 𝑒 to the three 𝑥 in both the numerator and denominator.
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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.
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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.
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Since there’s no 𝑥 dependency within our limit, our limit is simply equal to two-thirds.
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And this is the solution to the question.
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Many different limits give one of the indeterminate forms required to use L’Hopital’s rule.
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Let’s consider the following examples.
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Determine the limit as 𝑥 approaches one of negative 11 times the natural logarithm of 𝑥 over negative nine 𝑥 plus nine.
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Let’s start by trying to find this limit using direct substitution.
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We obtain negative 11 multiplied by the natural logarithm of one over negative nine plus nine.
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We will use the fact that the natural logarithm of one is equal to zero.
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And this gives us that our limit must be equal to zero over zero, which is undefined.
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However, this is the main condition which we must satisfy in order to use L’Hopital’s rule.
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This is L’Hopital’s rule.
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And since our limit is equal to zero over zero, we can see that we have satisfied the first condition.
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We’re taking the limit as 𝑥 approaches one.
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So we can say that 𝑎 is equal to one.
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And one is a real number.
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Therefore, we have satisfied the second condition.
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And this tells us that we’re able to use L’Hopital’s rule.
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We can say that 𝑓 of 𝑥 is equal to negative 11 timesed by the natural logarithm of 𝑥.
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And 𝑔 of 𝑥 is equal to negative nine 𝑥 plus nine.
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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.
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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.
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Now 𝑔 of 𝑥 is equal to negative nine 𝑥 plus nine, which is simply a polynomial.
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And so we can differentiate this using the power rule for differentiation, giving us that 𝑔 prime of 𝑥 is equal to negative nine.
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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.
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Which tells us that the limit as 𝑥 approaches one of negative 11 times the natural logarithm of 𝑥 over negative nine plus nine.
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Is equal to the limit as 𝑥 approaches one of negative 11 over negative nine 𝑥.
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We see that we have a factor of negative one in both the numerator and denominator.
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And so they can be cancelled, giving us the limit as 𝑥 approaches one of 11 over nine 𝑥.
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And here we can apply direct substitution, giving us 11 over nine times one.
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This gives us the solution that our limit must be equal to 11 over nine.
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In the next example, we’ll be seeing how we can use L’Hopital’s rule in order to find another result.
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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.
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Use L’Hopital’s rule to determine which is dominant as 𝑥 tends to infinity.
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The natural logarithm of 𝑥 or the square root of 𝑥.
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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.
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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.
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Or the limit as 𝑥 tends to infinity of the square root of 𝑥 over the natural logarithm of 𝑥 is equal to zero.
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Let’s start by considering the latter of these two options.
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We need to find the limit as 𝑥 tends to infinity of the square root of 𝑥 over the natural logarithm of 𝑥.
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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.
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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.
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And this is undefined.
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However, it does give us the main condition for using L’Hopital’s rule.
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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.
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Where 𝑎 is a real number, positive infinity, or negative infinity.
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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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Now since our limit is equal to positive infinity over positive infinity, we have satisfied the first condition.
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And we’re taking the limit as 𝑥 tends to positive infinity.
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Therefore, we’ve also satisfied the second condition.
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So we’re able to use L’Hopital’s rule.
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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.
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The square root of 𝑥 is also equal to 𝑥 to the power of a half.
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And so in order to find 𝑓 prime of 𝑥, we’ll use the power rule for differentiation.
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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.
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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.
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And so 𝑔 prime of 𝑥 is equal to one over 𝑥.
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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.
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Simplifying this, we obtain the limit as 𝑥 tends to infinity of 𝑥 over two timesed by 𝑥 to the power of one-half.
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Now we can cancel a factor of 𝑥 to the power of one-half from the top and bottom.
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We obtain the limit as 𝑥 tends to infinity of 𝑥 to the power of one-half over two.
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And the only 𝑥 term here has a positive power.
00:14:45.130 --> 00:14:46.840
And it’s in the numerator of the fraction.
00:14:47.180 --> 00:14:49.600
Therefore, this limit must be equal to infinity.
00:14:49.870 --> 00:14:51.870
And so it is not equal to zero.
00:14:52.220 --> 00:14:57.200
And we can conclude from this that the natural logarithm of 𝑥 does not dominate the square root of 𝑥.
00:14:57.420 --> 00:15:04.070
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.
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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.
00:15:11.230 --> 00:15:17.730
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.
00:15:17.830 --> 00:15:26.970
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.
00:15:27.310 --> 00:15:31.440
However, it allowed us to use L’Hopital’s rule, since these two conditions are satisfied.
00:15:31.860 --> 00:15:34.580
Our limit is equal to positive infinity over positive infinity.
00:15:34.830 --> 00:15:37.500
And we’re taking the limit as 𝑥 tends to positive infinity.
00:15:37.890 --> 00:15:42.670
Since we’re taking the limit of the reciprocal function, our 𝑓 and 𝑔 will be the other way around.
00:15:43.060 --> 00:15:47.850
Therefore, for our final line of working, we will simply be taking the limit of the reciprocal function.
00:15:48.150 --> 00:15:56.980
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.
00:15:57.550 --> 00:16:08.770
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.
00:16:09.080 --> 00:16:13.660
And that is the limit as 𝑥 tends to infinity of two over 𝑥 to the power of one-half.
00:16:14.180 --> 00:16:17.620
In this limit, our 𝑥 has a positive power of one-half.
00:16:17.930 --> 00:16:20.030
However, it’s in the denominator of the fraction.
00:16:20.450 --> 00:16:25.030
And therefore, if we take the limit as 𝑥 tends to infinity, it will be equal to zero.
00:16:25.620 --> 00:16:31.740
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.
00:16:32.170 --> 00:16:36.880
From this, we can conclude that the square root of 𝑥 dominates the natural logarithm of 𝑥.
00:16:38.490 --> 00:16:48.130
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.
00:16:48.700 --> 00:16:52.110
We’ve now seen a variety of examples involving L’Hopital’s rule.
00:16:52.480 --> 00:16:54.340
Let’s look at some key points of the video.
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Key Points.
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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.
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L’Hopital’s rule is as follows.
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Suppose the limit as 𝑥 approaches 𝑎 of 𝑓 of 𝑥 over 𝑔 of 𝑥 equals zero over zero.
00:17:13.040 --> 00:17:20.140
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.
00:17:20.380 --> 00:17:24.420
Where 𝑎 can be any real number, positive infinity, or negative infinity.
00:17:24.910 --> 00:17:32.770
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.