We're asked to evaluate the limit.
We have the limit of the complex fraction
as b approaches two.
Notice here if we try to perform direct substitution
to determine the limit, we have 1/2 minus 1/2,
which is zero in the numerator.
In the denominator, we have two minus two,
which is also zero.
This gives us the indeterminate form of zero divided by zero
which indicates we cannot determine the limit
by performing direct substitution,
but that does not mean the limit does not exist.
We can often perform algebraic techniques
to change the form of the function
we're trying to find the limit of,
so that we can then find the limit
by performing direct substitution.
In this case, the only thing we can possibly do
to try to simplify the complex fraction
would be to subtract the fractions in the numerator.
And that's what we'll do.
This actually falls under the case of case three
where we find the least common denominator.
So I'm going to show two ways we can simplify
this complex fraction in order to determine the limit
by performing direct substitution.
So let's go ahead and rewrite the complex fraction below.
We have one divided by b minus 1/2
divided by the quantity b minus two.
So now we'll begin by subtracting these two fractions
by obtaining the least common denominator,
which is two b.
So we'll multiply the fraction one over b by two
in the numerator and denominator.
We'll multiply the fraction of 1/2 by b
in the numerator and denominator.
And then we'll go ahead and subtract.
Notice now we have a common denominator of two b
in the numerator subtracting,
we have two minus b in the numerator.
And remember a fraction represents division,
so we can write this as two minus b all over two b
divided by b minus two.
If we went b minus two over one,
now we'll write this as a product
dividing by b minus two is equivalent
to multiplying by the reciprocal of one over b minus two.
So this gives us two minus b over two b times
one over the quantity b minus two.
Now, if we look at two minus b and b minus two,
these two expressions are opposites.
Notice for two minus b, the two is positive
and the b is negative.
And for b minus two, the b is positive
and the two is negative.
So we're going to go ahead and factor out our negative one
from two minus b.
This will give us negative one times
the quantity negative two plus b.
And we can write negative two plus b as b minus two.
So let's rewrite this as negative one times
the quantity b minus two all over two b,
times one over the quantity b minus two.
And now we can see the common factor
of b minus two between the numerator and denominator.
b minus two divided by itself simplifies to one.
The given complex fraction simplifies
to negative one over two b.
Which means you can rewrite the visual limit
as the limit of negative one divided by two b
as b approaches two.
So I showed some extra work here.
I just want to show a shorter way by multiplying
the numerator and the denominator of the complex fraction
by the LCD of two b.
So we can now show this one other way,
starting with the original expression again.
Again, recognizing the LCD of the numerator
is going to be two b, we would multiply the numerator
by two b and the denominator by two b, and then simplify.
In the numerator, one over b times two b is two.
And then minus two b times 1/2 is b.
This is still divided by, we're going to leave the denominator
in factored form as two b times the quantity b minus two.
And again from here we will factor out negative one
from the numerator, which gives us negative one
times the quantity, we know from above,
we're going to have negative two plus b
in the parentheses, we can write as b minus two.
In this form again, we can see we have a common factor
of b minus two between the numerator and denominator.
And of course, we get the same simplified expression
of negative one over two b.
Either method works for simplifying the complex fraction.
The whole point is we can now find the limit
by performing direct substitution.
Just negative one divided by two b
is continuous everywhere except at b equals zero,
and we're approaching b equals two.
Evaluating the limit by performing direct substitution,
we now have negative one divided by the product
of two and two, which gives us negative 1/4 the limit.
Let's also verify this limit graphically as well
using a table of values.
Again, for the limit to exist,
we must be approaching the same function value
as you approach two from the right and the left
or from values less than two and greater than two.
Looking at the table below,
notice as we approach two from values less than two,
we can see that the function values are approaching
negative 0.25 or negative 1/4.
And as we approach two from values greater than two,
again, we can see, we are approaching
the same function value
of negative 0.25 or negative 1/4.
And graphically, if we draw a vertical line at b equals two,
which would be here,
as we approach two from the left and from the right,
we can see graphically, we are approaching
the same function value.
It's a little hard to tell,
but we are approaching the y value
or function value of negative 1/4,
which does verify the limit is negative 1/4
I hope you found this helpful.