Most of what we do early on when we first learn about
calculus is to use limits.
We use limits to figure out derivatives of functions.
In fact, the definition of a derivative uses
the notion of a limit.
It's a slope around the point as we take the limit of
points closer and closer to the point in question.
And you've seen that many, many, many times over.
In this video I guess we're going to do it in the
opposite direction.
We're going to use derivatives to figure out limits.
And in particular, limits that end up in indeterminate form.
And when I say by indeterminate form I mean that when we just
take the limit as it is, we end up with something like 0/0, or
infinity over infinity, or negative infinity over
infinity, or maybe negative infinity over negative
infinity, or positive infinity over negative infinity.
All of these are indeterminate, undefined forms.
And to do that we're going to use l'Hopital's rule.
And in this video I'm just going to show you what
l'Hoptial's rule says and how to apply it because it's fairly
straightforward, and it's actually a very useful tool
sometimes if you're in some type of a math competition and
they ask you to find a difficult limit that when you
just plug the numbers in you get something like this.
L'Hopital's rule is normally what they are testing you for.
And in a future video I might prove it, but that gets a
little bit more involved.
The application is actually reasonably straightforward.
So what l'Hopital's rule tells us that if we have-- and I'll
do it in abstract form first, but I think when I show you
the example it will all be made clear.
That if the limit as x approaches c of f of x is equal to 0, and
the limit as x approaches c of g of x is equal to 0, and-- and
this is another and-- and the limit as x approaches c of f
prime of x over g prime of x exists and it equals L.
then-- so all of these conditions have to be met.
This is the indeterminate form of 0/0, so this
is the first case.
Then we can say that the limit as x approaches c of
f of x over g of x is also going to be equal to L.
So this might seem a little bit bizarre to you right now, and
I'm actually going to write the other case, and then
I'll do an example.
We'll do multiple examples and the examples are going
to make it all clear.
So this is the first case and the example we're going to
do is actually going to be an example of this case.
Now the other case is if the limit as x approaches c of f of
x is equal to positive or negative infinity, and the
limit as x approaches c of g of x is equal to positive or
negative infinity, and the limit of I guess you could say
the quotient of the derivatives exists, and the limit as x
approaches c of f prime of x over g prime of x
is equal to L.
Then we can make this same statement again.
Let me just copy that out.
Edit, copy, and then let me paste it.
So in either of these two situations just to kind of make
sure you understand what you're looking at, this is the
situation where if you just tried to evaluate this limit
right here you're going to get f of c, which is 0.
Or the limit as x approaches c of f of x over the limit as
x approaches c of g of x.
That's going to give you 0/0.
And so you say, hey, I don't know what that limit is?
But this says, well, look.
If this limit exists, I could take the derivative of each
of these functions and then try to evaluate that limit.
And if I get a number, if that exists, then they're going
to be the same limit.
This is a situation where when we take the limit we get
infinity over infinity, or negative infinity or positive
infinity over positive or negative infinity.
So these are the two indeterminate forms.
And to make it all clear let me just show you an example
because I think this will make things a lot more clear.
So let's say we are trying to find the limit-- I'll
do this in a new color.
Let me do it in this purplish color.
Let's say we wanted to find the limit as x approaches
0 of sine of x over x.
Now if we just view this, if we just try to evaluate it at 0 or
take the limit as we approach 0 in each of these functions,
we're going to get something that looks like 0/0.
Sine of 0 is 0.
Or the limit as x approaches 0 of sine of x is 0.
And obviously, as x approaches 0 of x, that's also
going to be 0.
So this is our indeterminate form.
And if you want to think about it, this is our f of x, that
f of x right there is the sine of x.
And our g of x, this g of x right there for this
first case, is the x.
g of x is equal to x and f of x is equal to sine of x.
And notice, well, we definitely know that this meets the
first two constraints.
The limit as x, and in this case, c is 0.
The limit as x approaches 0 of sine of sine of x is 0, and
the limit as x approaches 0 of x is also equal to 0.
So we get our indeterminate form.
So let's see, at least, whether this limit even exists.
If we take the derivative of f of x and we put that over the
derivative of g of x, and take the limit as x approaches 0
in this case, that's our c.
Let's see if this limit exists.
So I'll do that in the blue.
So let me write the derivatives of the two functions.
So f prime of x.
If f of x is sine of x, what's f prime of x?
Well, it's just cosine of x.
You've learned that many times.
And if g of x is x, what is g prime of x?
That's super easy.
The derivative of x is just 1.
Let's try to take the limit as x approaches 0 of f prime of x
over g prime of x-- over their derivatives.
So that's going to be the limit as x approaches 0
of cosine of x over 1.
I wrote that 1 a little strange.
And this is pretty straightforward.
What is this going to be?
Well, as x approaches 0 of cosine of x, that's
going to be equal to 1.
And obviously, the limit as x approaches 0 of 1, that's
also going to be equal to 1.
So in this situation we just saw that the limit as x
approaches-- our c in this case is 0.
As x approaches 0 of f prime of x over g prime
of x is equal to 1.
This limit exists and it equals 1, so we've met
all of the conditions.
This is the case we're dealing with.
Limit as x approaches 0 of sine of x is equal to 0.
Limit as z approaches 0 of x is also equal to 0.
The limit of the derivative of sine of x over the derivative
of x, which is cosine of x over 1-- we found this
to be equal to 1.
All of these top conditions are met, so then we know
this must be the case.
That the limit as x approaches 0 of sine of x over x
must be equal to 1.
It must be the same limit as this value right here where we
take the derivative of the f of x and of the g of x.
I'll do more examples in the next few videos and I think
it'll make it a lot more concrete.