We want to find the limit as x approaches one
of the quantity x to the third minus one,
divided by the quantity x squared minus one.
But notice how as x approaches positive one,
both the numerator and the denominator approach zero
and therefore we say this limit has the indeterminate form
of zero divided by zero.
But because of this indeterminate form,
we can imply L'Hopital's Rule, to help us determine
this limit.
L'Hopital's Rule, or Bernoulli's Rule,
tells us that if a limit has one of these indeterminate
forms listed here, the limit as x approaches c
of f of x divided by g of x, is equal to
the limit as x approaches c of f prime of x,
divided by g prime of x.
So if this limit fits one of these forms here,
we can find the derivative of f, the derivative of g
and see if we can determine this limit.
So going back to our example, notice how we might
try to factor this, but it's going to apply
L'Hopital's Rule.
Which means this limit is equal to the limit
as x approaches positive one, of the derivative of
three x cubed minus one, divided by the derivative of
x squared minus one.
Well the derivative of x to the third would be
three x squared.
The derivative of one would be zero, and
the derivative of x squared would be two x,
derivative of one is zero.
Notice in this form, we could simplify out a common factor
of x, but there's no need to.
Notice as x approaches one, the numerator approaches three,
the denominator approaches two, and therefore
this limit is equal to three halves.
Let's see if we can verify this by looking at the graph
of the function.
Notice here's where x equals one.
And as we approach positive one, from the right
or from the positive side, and as we approach
positive one from the negative side or left side,
we're approaching the same function value
which we can see would be three halves, verifying our limit.
Looking at our second example, we have the limit
as x approaches zero, of natural log of the quantity
two x plus one, divided by five x.
Well notice, as x approaches zero, this approaches
natural log one, which is equal to zero, and as x
approaches zero, five x approaches zero.
So once again, this limit is in the indeterminate form
of zero divided by zero, so let's apply L'Hopital's Rule
to help us find this limit.
The derivative of natural log, the quantity two x
plus one, where the quantity two x plus one would be
the inner function or u, so the derivative is going to be
one over u times u prime, or one over the quantity
two x plus one, times the derivative of two x
plus one, which would be two.
Divided by the derivative of five x,
which would be five.
Now let's go and simplify this.
We have two divided by the quantity two x plus one,
divided by five, which is the same as multiplying
by the reciprocal of one fifth.
So times one fifth.
Which means we have the limit as x approaches zero
of two, divided by five, times the quantity
two x plus one.
So notice as x approaches zero, the numerator is
still two.
Our denominator would approach five times the quantity
two times zero, plus one, which would just be two
divided by five, times one, or two fifths.
So our limit is equal to two fifths.
Again, let's try to verify this by looking at the graph
of the given function.
Here's where x equals zero.
As we approach x equals zero from the right
or from the positive side, we're approaching in
this direction here.
As we approach zero from the left or the negative side,
we're approaching in this direction here.
We are approaching the same value from both sides.
The function value here or the Y value,
does look like it's approximately two fifths or 0.4.
Therefore, this does verify our limit.
I hope this was helpful.