Introduction to l'Hôpital's rule - Mathematics Video Lecture - Engineering Mathematics

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 roaches
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.