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Most of what we do early on when we first learn about

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calculus is to use limits.

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We use limits to figure out derivatives of functions.

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In fact, the definition of a derivative uses

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the notion of a limit.

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It's a slope around the point as we take the limit of

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points closer and closer to the point in question.

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And you've seen that many, many, many times over.

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In this video I guess we're going to do it in the

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opposite direction.

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We're going to use derivatives to figure out limits.

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And in particular, limits that end up in indeterminate form.

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And when I say by indeterminate form I mean that when we just

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take the limit as it is, we end up with something like 0/0, or

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infinity over infinity, or negative infinity over

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infinity, or maybe negative infinity over negative

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infinity, or positive infinity over negative infinity.

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All of these are indeterminate, undefined forms.

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And to do that we're going to use l'Hopital's rule.

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And in this video I'm just going to show you what

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l'Hoptial's rule says and how to apply it because it's fairly

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straightforward, and it's actually a very useful tool

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sometimes if you're in some type of a math competition and

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they ask you to find a difficult limit that when you

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just plug the numbers in you get something like this.

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L'Hopital's rule is normally what they are testing you for.

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And in a future video I might prove it, but that gets a

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little bit more involved.

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The application is actually reasonably straightforward.

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So what l'Hopital's rule tells us that if we have-- and I'll

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do it in abstract form first, but I think when I show you

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the example it will all be made clear.

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That if the limit as x roaches c of f of x is equal to 0, and

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the limit as x approaches c of g of x is equal to 0, and-- and

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this is another and-- and the limit as x approaches c of f

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prime of x over g prime of x exists and it equals L.

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then-- so all of these conditions have to be met.

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This is the indeterminate form of 0/0, so this

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is the first case.

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Then we can say that the limit as x approaches c of

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f of x over g of x is also going to be equal to L.

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So this might seem a little bit bizarre to you right now, and

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I'm actually going to write the other case, and then

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I'll do an example.

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We'll do multiple examples and the examples are going

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to make it all clear.

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So this is the first case and the example we're going to

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do is actually going to be an example of this case.

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Now the other case is if the limit as x approaches c of f of

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x is equal to positive or negative infinity, and the

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limit as x approaches c of g of x is equal to positive or

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negative infinity, and the limit of I guess you could say

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the quotient of the derivatives exists, and the limit as x

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approaches c of f prime of x over g prime of x

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is equal to L.

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Then we can make this same statement again.

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Let me just copy that out.

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Edit, copy, and then let me paste it.

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So in either of these two situations just to kind of make

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sure you understand what you're looking at, this is the

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situation where if you just tried to evaluate this limit

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right here you're going to get f of c, which is 0.

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Or the limit as x approaches c of f of x over the limit as

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x approaches c of g of x.

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That's going to give you 0/0.

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And so you say, hey, I don't know what that limit is?

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But this says, well, look.

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If this limit exists, I could take the derivative of each

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of these functions and then try to evaluate that limit.

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And if I get a number, if that exists, then they're going

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to be the same limit.

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This is a situation where when we take the limit we get

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infinity over infinity, or negative infinity or positive

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infinity over positive or negative infinity.

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So these are the two indeterminate forms.

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And to make it all clear let me just show you an example

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because I think this will make things a lot more clear.

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So let's say we are trying to find the limit-- I'll

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do this in a new color.

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Let me do it in this purplish color.

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Let's say we wanted to find the limit as x approaches

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0 of sine of x over x.

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Now if we just view this, if we just try to evaluate it at 0 or

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take the limit as we approach 0 in each of these functions,

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we're going to get something that looks like 0/0.

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Sine of 0 is 0.

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Or the limit as x approaches 0 of sine of x is 0.

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And obviously, as x approaches 0 of x, that's also

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going to be 0.

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So this is our indeterminate form.

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And if you want to think about it, this is our f of x, that

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f of x right there is the sine of x.

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And our g of x, this g of x right there for this

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first case, is the x.

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g of x is equal to x and f of x is equal to sine of x.

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And notice, well, we definitely know that this meets the

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first two constraints.

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The limit as x, and in this case, c is 0.

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The limit as x approaches 0 of sine of sine of x is 0, and

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the limit as x approaches 0 of x is also equal to 0.

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So we get our indeterminate form.

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So let's see, at least, whether this limit even exists.

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If we take the derivative of f of x and we put that over the

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derivative of g of x, and take the limit as x approaches 0

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in this case, that's our c.

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Let's see if this limit exists.

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So I'll do that in the blue.

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So let me write the derivatives of the two functions.

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So f prime of x.

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If f of x is sine of x, what's f prime of x?

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Well, it's just cosine of x.

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You've learned that many times.

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And if g of x is x, what is g prime of x?

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That's super easy.

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The derivative of x is just 1.

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Let's try to take the limit as x approaches 0 of f prime of x

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over g prime of x-- over their derivatives.

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So that's going to be the limit as x approaches 0

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of cosine of x over 1.

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I wrote that 1 a little strange.

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And this is pretty straightforward.

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What is this going to be?

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Well, as x approaches 0 of cosine of x, that's

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going to be equal to 1.

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And obviously, the limit as x approaches 0 of 1, that's

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also going to be equal to 1.

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So in this situation we just saw that the limit as x

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approaches-- our c in this case is 0.

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As x approaches 0 of f prime of x over g prime

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of x is equal to 1.

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This limit exists and it equals 1, so we've met

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all of the conditions.

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This is the case we're dealing with.

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Limit as x approaches 0 of sine of x is equal to 0.

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Limit as z approaches 0 of x is also equal to 0.

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The limit of the derivative of sine of x over the derivative

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of x, which is cosine of x over 1-- we found this

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to be equal to 1.

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All of these top conditions are met, so then we know

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this must be the case.

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That the limit as x approaches 0 of sine of x over x

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must be equal to 1.

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It must be the same limit as this value right here where we

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take the derivative of the f of x and of the g of x.

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I'll do more examples in the next few videos and I think

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it'll make it a lot more concrete.

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