Welcome to a second video on L'Hopitals Rule.
Let's go and do a quick review before we look at our
additional examples.
L'Hopitals Rule uses derivatives to evaluate limits
involving the following indeterminant forms.
Even if the limit is not in this form, we can often
manipulate the function to obtain a form or we can apply
L'Hopitals Rule.
And this is what this video focuses on.
L'Hopitals Rule states that the limit of a quotient,
as we see here, is equal to the limit of the quotient
of the derivatives, under special conditions as we see here.
The special conditions are described here, you may want
to pause the video and read these, we did discuss them
in the first video.
Let's go and look at our additional examples.
Sometimes logarithms can help rewrite limits before using
L'Hopitals Rule.
The limit is x to the power infinity of one plus one over x
raised to the power of x.
Our approach is one to the power of infinity,
which is an indeterminant form but it's not a form
where we can apply L'Hopitals Rule.
Though we can use logarithms to help, so if we set
this limit equal to y.
And then take the natural log of both sides, we'd have
natural log y equals natural log of the limit.
The natural log of this limit is equal to the limit
of the natural log.
So we have natural log y equals the limit natural log
one plus one over x to the power of x.
And now if we apply the power property of log we can move
this x to x times the natural log.
So let's go ahead and do that.
Now take a look and see what happens to this as x approaches
infinity, this goes to infinity, and this approaches
natural log one which is zero.
One infinity times zero is still indeterminant,
It's still not in the form we need to apply L'Hopitals Rule.
So notice that we're multiplying by x.
Instead of multiplying by x we could divide
by the reciprocal, let's go and see what that does for us.
Again we normally wouldn't do this, but dividing by one
over x is the same as multiplying by the reciprocal
or times x over one, which is the same as multiplying by x.
Now the reason we want to do that here is, as x approaches
infinity, the numerator approaches natural log one
which is zero.
And one over x also approaches zero, so now we have
the indeterminant form of zero over zero, so in this form
we can apply L'Hopitals Rule.
Before we find these derivatives, let's just remember
that one over x is equal to x to the power of negative one.
So looking at the numerator, we'll have to apply
the chain rule here where u is equal to one plus one over x.
So the derivative of natural log u is one over u times
u prime, or one over one plus one over x times u prime.
Well the derivative of one is zero, derivative of one
over x, which remember is x to the power of negative one
would be negative one x to the negative two.
Notice in our denominator we have x to the power
of negative one as well, so that derivative would be
the same, negative one over x to the negative two.
And now we can see this simplifies nicely, this over this
would equal one.
So we have natural log y equals the limit,
as x approaches infinity of one over one plus one over x.
Well one over x goes to zero as x approaches infinity
so we just have one over one or one.
Now be careful we're not done here.
We're actually trying to find this limit which is equal to y
and we have found that natural log y is equal to one.
So we need to solve this equation for y.
Remember natural log is base e so we have e to the one
is equal to y.
So our original limit,
is equal to e.
And this is quite interesting because we just found
the definition for the number e.
Let's go and take a look at another one.
Notice here we have the limit as x approaches zero
of a difference and both of these fractions as x approaches
zero approach infinity.
So we have the form of infinity minus infinity
which again is indeterminant but does not fit the form
for L'Hopitals Rule.
So let's go ahead and add these two fractions together
and then see what we have.
So this first fraction is missing the factor of e
to the x minus one.
And this fraction is missing the factor of x.
So our common denominator is x times e to the x minus one.
Now I'm going to go ahead and multiply that out.
We'd have x, e to the x, minus x.
And the numerator would be e to the x minus one minus x.
Let's see what happens when x approaches zero.
We'd have, this approaches one, so we'd have one minus one
minus zero, so our numerator's approaching zero
and our denominator is also going to approach zero
as x approaches zero.
So in this form we can apply L'Hopitals Rule.
Let's go ahead and find the derivative of the numerator
and denominator.
Derivative of e to the x is e to the x, derivative of one
is zero, derivative of x is equal to one,
so we have minus one.
Now here we're going to have to apply the product rule.
It's x times e to the x, so we'll have the first times
derivative of the second, that we be x e to the x
plus the second e to the x, times the derivative of x
which is one.
So we have plus e to the x the derivative of x would be one.
Now if we try to perform substitution in this form we'd have
one minus one that would be zero in the numerator
and then in our denominator we would have zero plus one
minus one which is still zero.
So this limit is still in the indeterminant form of zero
over zero, but remember we can apply L'Hopitals Rule
more than once.
So let's go ahead and find the derivatives again.
Derivative of e to the x is e to the x, derivative of one
would be zero.
Here we have to apply that product rule again so we have
the first times derivative of the second so x e to the x
plus the second which e to the x times zero of x
which is one, plus derivative of e to the x
that's e to the x derivative of one is zero.
Now let's see what we have.
If we try to perform direct substitution we'd have
e to the zero that would be one, in the denominator we'd
have zero plus one, plus one, so this limit is equal
to one half.
I hope you found these additional examples helpful.