WEBVTT mathematics/ap-calculus-ab/hovasapian
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Hello, and welcome back to www.educator.com, and welcome back to AP Calculus.
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Today, we are going to talk about L’Hospital’s rule.
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Today's lesson is going to be mostly theory.
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There are some examples in here just to bolster the theory.
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But it is the next lesson that we are actually going to be doing most of the example problems.
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Let us dive right on in.
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I will stick with black today.
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Recall this limit which we did a while back, it is the limit as x goes to 0 of sin x/ x.
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We had some procedure that we did, in order to actually come up with this limit.
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When we plug in 0, the way we normally do, into this, we get 0/0.
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That was a bit of a problem.
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We have this alternate procedure, in order to come up with our particular limit that we go for this.
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How about this one?
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How about the limit as x goes to infinity of the nat-log (x)/ x – 2.
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As x goes to infinity, the numerator goes to infinity and the denominator also goes to infinity.
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What we end up with is this thing, infinity/ infinity.
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0/0 and infinity/ infinity, whenever we want to cross things like this when we are taking limits,
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these are called indeterminate forms.
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They are called indeterminate forms, in other words, we do not know.
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One of them is going to be moving to 0 faster than the other.
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One of them is going to be going to infinity faster than the other.
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Does the positive infinity win, does the other infinity win?
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Did they meet some place in the middle, we do not know.
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It is indeterminate, indeterminate forms.
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There is a general way, there is a beautiful general procedure.
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There is a beautiful general way to handle these, when you get either of these indeterminate forms.
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If you ever get infinity/ infinity or 0/0, this is what you do.
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This is called L’Hospital’s rule.
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Here is the theorem.
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Now that I’m actually going to be reading of the theorem, I think I will go to blue.
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We will let f(x) and g(x) be differentiable and g’(x) not equal to 0, on an open interval.
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On an open interval a, that contains the point a.
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You should know that differentiability at a is not necessary.
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Differentiability at a is not necessary.
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It may or may not be differentiable there, because we are dealing with limits.
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Differentiability at a is not necessary, it could go either way on that one.
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If the limit as x approaches a(f) is equal to 0 and the limit as x approaches a(g) is equal to 0,
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then the limit as x approaches a(f/g) is equal to the limit as x approaches a(f’/f’).
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What this is telling me is that, if I ever come up with an indeterminate form where I end up with 0/0,
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all I have to do is take the derivative of the numerator, take the derivative of the denominator,
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and then take the limit again.
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If it happens again, I do it again, I do it again, until I end up with a limit.
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A finite limit, a number, 0 or infinity, that is it.
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You just keep differentiating the numerator and the denominator.
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This is not quotient rule.
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You differentiate the numerator, you differentiate the denominator.
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You come up with f’/g’, and then you take the limit again.
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This is a really beautiful thing.
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The other part of this, the other indeterminate form involving infinity says,
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if the limit as x approaches a(f) = + or –infinity and the limit as x approaches a(g) = + or –infinity.
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In other words, if you end up with the indeterminate form infinity/ infinity, + or -, you can go ahead and apply this.
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Then, the limit as x approaches a(f/g), the original function is equal to the limit as x approaches a(f’/g’), the same thing.
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If you ever end up with an indeterminate form upon taking limits, when you plug the a into whatever it is,
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infinity/ infinity is what you get, you can just take the derivative of the top and bottom, take the limit again.
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Keep going until you get your answer.
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Very nice, you will use this a lot in all of your work, no matter what field you go into.
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In other words, if you get 0/0 or infinity/infinity then you can take f’ and g’, and evaluate the limit again.
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Keep going until you arrived at a viable limit.
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In other words, a number, or 0, or infinity, there you go.
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Again, this is not the quotient rule.
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You have f/g, if you end up taking the limit of that, plugging in the a, you get 0/0 or you end up with infinity/ infinity,
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you take the derivative of the numerator f’/ the derivative of the denominator, and take the limit again.
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This is not quotient rule, be very careful with this.
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Let us do an example, evaluate the following limit using L’Hospital’s rule.
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Putting 2 into this, putting 2 into f and g.
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F is the top and g is the bottom.
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We get ln of ½ × 2/ 2 – 2.
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This is 0/0, we have an indeterminate form, we can apply L’Hospital’s rule.
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We differentiate the numerator.
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When we differentiate the numerator, we get 1/ 1/2x × ½/ the derivative of x - 2 is 1.
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This cancels and we are left with 1/x.
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The limit as x approaches 2 of 1/x = ½, that is our limit of the original.
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That is it, nice and easy.
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Derivative of the numerator, derivative of the denominator, take the limit again.
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If you end up with 0/0 again or infinity/ infinity, take the derivative of the numerator/ the derivative of the denominator, take the limit again.
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Keep going until you get a finite limit, that is when you stop.
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Let us do another example, evaluate the following limit using L’Hospital’s rule.
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The limit as x approaches π/2 from the positive side of cos x/1 - sin x.
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When I put π/2 in here, this is going to give me 0.
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When I put π/2 in here, it is going to give me 1 – 1, it is going to give me 0.
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0/0 this is indeterminate, I can apply L’Hospital’s rule.
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I take the derivative of cos x is – sin x.
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I take the derivative of 1 - sin x which is - cos x.
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I take the limit again, as x approaches π/2 from the positive.
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When I put π/2 in for here, -sin(π/2) is – 1.
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This is 0, this goes to –infinity.
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As x approaches π/2, -cos x approaches 0.
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As it approaches 0, this thing goes to –infinity.
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-infinity is a perfectly viable limit.
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Let us talk about indeterminate products.
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We have indeterminate products.
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If the limit as x approaches a of f(x) × g(x), if you have two functions that are multiplied together.
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If it gives 0 × a + or –infinity, this is another indeterminate form.
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0 × infinity, infinity × 0, is indeterminate.
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L’Hospital’s rule applies when you have 0/0 or infinity/infinity.
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What you are going to do is you are going to take this f(x) and g(x),
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rewrite it in such a way that you get 0/0 or infinity/infinity.
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And then, apply L’Hospital’s rule, that is it.
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If you ever take the limit of a product and you get 0 × infinity or infinity × 0, this is indeterminate.
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Handle this by rewriting f × g, you can write it as f/ 1/g or you can rewrite it as g/ 1/f.
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In other words, drop one of them down as a reciprocal into the denominator which will give you 0/0 or infinity/infinity.
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And then, apply L’Hospital’s rule and take the limit.
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Apply L’Hospital’s rule, let us do an example.
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We have evaluate the following limit using L’Hospital’s rule.
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The limit as x approaches 0 of x ln x, as x approaches 0 from the positive side.
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Because again, the negative numbers are not in the domain of ln x.
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When we plug these in, we are going to get 0 ×, the ln of 0, as x approaches 0, ln x becomes –infinity.
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This is indeterminate.
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Because it is indeterminate, we can manipulate it and then apply L’Hospital’s rule.
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Let us rewrite x ln x, it is equal to ln x/ 1/x.
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When I take the limit of this, as x goes to 0, I’m going to get -infinity/infinity.
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This is an indeterminate form, now we can apply L’Hospital’s rule because L’Hospital’s rule has 0/0 or infinity/infinity.
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L’Hospital’s rule gives us, now that we have this, we are going to take the derivative of the top and the derivative of the bottom.
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The derivative of the top is 1/x, the derivative of the bottom is -1/ x².
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This is equal to –x.
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When we take the limit as x approaches 0 of – x, we get 0.
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There we go, we had a product, we rearrange that product to give us infinity/ infinity.
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We treat L’Hospital’s rule, we take the derivative of the top and the derivative of the bottom.
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Take the limit again of what we get.
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We could have also done x × the nat-log of x, we could have dropped the nat-log into the denominator as x/1/ ln x.
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When we differentiate this, it is going to give us 0/0.
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It is not a problem, when we take the limit.
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But when you differentiate this, you are going to get something that is more complex than this.
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That is it, there is no way of knowing beforehand which one you are going to drop into the numerator or denominator.
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I’m sorry, drop into the denominator.
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You just sort of try one.
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It might end up being really easy, like this was.
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It might end up being more complex, so you try the other one.
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That is it, you just try and you come up with something.
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Let us do indeterminate differences.
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If we evaluate the limit as x approaches a of f(x) – g(x), and get infinity – infinity, this is an indeterminate difference.
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This is indeterminate.
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I do not know which one is going to infinity faster.
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I do not know which is going to dominate, the positive infinity or the negative infinity,
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or they might meet someplace in between at a finite number, we do not know.
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It is definitely indeterminate.
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The way we handle this, by trying to convert it into something which is 0/0 or infinity/infinity,
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and apply L’Hospital’s rule, the L’Hospital’s theorem.
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We handle this by trying to convert this difference, this difference, to a quotient, in order to get either 0/0 or infinity/infinity.
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It is these two indeterminate forms that allow us to apply L’Hospital’s rule.
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Let us see what we can do.
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Evaluate the following limit using L’Hospital’s rule.
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X goes to infinity of x – ln x.
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As we take this, we definitely get infinity – infinity.
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This is definitely an indeterminate difference.
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Wait a minute, I’m actually looking at a different a problem here.
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Look at that, I did not even notice.
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I have one problem written here and I have another problem written on my sheet.
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I'm actually going to rewrite this problem and give you a new one.
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We are not going to do that one, we are going to do this one.
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X goes to 0 of csc x – cot(x).
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Once again, when we put this in, as x approaches 0, we are going to end up with infinity – infinity.
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We are going to rewrite csc x – cot x.
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We are going to write it in terms of basic functions.
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This is going to be 1/ sin x - cos x/ sin x which = 1 - cos x/ sin x.
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When I take the limit as x approaches 0 of this one, which is the exact same function, I have just rewritten it, I get 0/0.
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Now I have an indeterminate form which allows me to apply L’Hospital’s rule.
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L’Hospital’s rule applies, therefore, I take the derivative of this and the derivative of that.
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The derivative of that is going to be sin x, the numerator.
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The derivative of the denominator is going to be cos x.
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I’m going to take the limit again, as x approaches to 0.
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This, as x approaches 0, sin x approaches 0.
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As x approaches 0, cos x approaches 1.
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This is equal to 0, that is my limit.
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There is probably 5 or 6 different ways that you can actually do this.
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There is no law that says you have to use L’Hospital’s rule.
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If you can figure out a way that allows you to do this with algebraic manipulation,
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playing around with this, playing around with that, you are more than welcome to do so.
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This is just one more tool in your toolbox, that is all this is.
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You should always use all the powers at your disposal, in order to solve a given problem.
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You do not feel like you have to be constrained by a certain rule.
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Let us move on to our final topic here which is going to be indeterminate powers.
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Whenever we are faced with a limit as x approaches a of f(x) ⁺g(x).
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When this gives 0⁰ or if it gives infinity⁰, or if it gives 1 ⁺infinity, these are indeterminate.
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These are indeterminate, we need to manipulate these to make them something where we can apply L’Hospital’s rule.
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We deal with these as follows, one way of doing it is by taking the nat-log.
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When you do this, when you take the nat-log of this function, the nat-log of f(x) ⁺g(x), you end up with g(x) × the nat-log of f(x).
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You have something that looks like a product.
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You create the product, you have an indeterminate product,
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and then you convert that into something that looks like L’Hospital’s rule, 0/0, infinity/infinity.
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And then, you go ahead and apply L’Hospital’s rule.
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The other way of doing it is by writing the function as an exponential.
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Writing the expression as an exponential.
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In other words, e ⁺ln of f(x) ⁺g(x) which is going to end up equaling e ⁺g(x) × ln of f(x), which we can deal with appropriately.
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Let us go ahead and do our final example.
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Again, these are just examples to bolster our theory.
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In the next lesson, we are going to do a lot more examples.
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Evaluate the following limit using L’Hospital’s rule.
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The limit as x approaches 0 of 1 + sin x ⁺cot x.
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As x approaches 0, sin(x) is going to go to 0.
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This is going to equal 1.
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As x approaches 0, cot x goes to infinity.
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This is 1 ⁺infinity, this definitely qualifies as an indeterminate power.
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Here our function is y = 1 + sin x ⁺cot x.
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When we take the natlog of both sides, we are going to get the natlog of y = the natlog of this
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which is going equal cot(x) × the natlog of 1 + sin x.
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When I take the limit of this, when I take x going to 0, I'm going to get infinity × 0.
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This is an indeterminate product, so far so good, we are working out.
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How we deal with an indeterminate product?
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We drop one of them down into the denominator.
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I’m going to write this as ln y = ln of 1 + sin x/ 1/ cot(x).
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This is nothing more than the ln of 1 + sin x, 1/cot is the tan(x).
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When I take the limit as x approaches 0, I'm going to get the natlog of 1 is 0.
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This is going to be 0/0.
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Now that I have an indeterminate form that matches L’Hospital’s rule, I can apply L’Hospital’s rule.
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In other words, I’m going to take the derivative of the top, the derivative of the bottom.
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The derivative of the top, the derivative of the logarithm is going to end up being, when I take the derivative,
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I’m going to skip a couple of steps, I hope you do not mind.
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I'm going to get cos(x) divided by 1 + sin(x).
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The derivative of the bottom is going to be sec² x.
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I’m going to go ahead and take the limit of that, as x approaches 0.
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I'm going to get 1/1 for the top and I'm going to get 1 for the bottom, which = 1.
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That is my final answer.
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The limit, but notice, I found the limit of this thing which is actually the natlog of y.
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I did not find the limit of y itself.
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I have to take that extra step and bring it back.
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We ended up taking the natlog to find the limit.
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It is the limit as x approaches 0 +, of the natlog of y, that is what = 1.
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Let us go ahead and bring it back to the actual limit itself.
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Let us just recap, we wanted the limit as x approaches 0 of y.
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Y is equal to e ⁺ln y.
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The limit as x approaches 0 ^+ of y = the limit as x approaches 0, e ⁺ln y which equals e¹.
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We found the limit of that which is equal to e.
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There we go, we found our limit.
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We had a power, we took the logarithm of it.
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The logarithm gave us a product.
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We took that product, we turn it into something which is 0/0.
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We apply L’Hospital’s rule, we found a limit of the natlog of y because of that first process to simplify it.
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Now we just reverse it.
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If you take the logarithm of something, if you want to go back, just exponentiate it.
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We do not really need to go through this.
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This is just going to be e¹.
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Once again, we will continue on with more examples, exclusively examples applying L’Hospital’s rule in the next lesson.
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Thank you so much for joining us here at www.educator.com.
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We will see you next time, bye.