We want to find the limit
of the rational function
as X approaches five.
If the limit does exist,
the function must be approaching the same function value
from the left and right of X equals five.
If the function is continuous
at and around positive five,
we can determine the limit
by performing direct substitution.
In this case, though, if we substitute five for X,
we get the indeterminate form of zero divided by zero.
This does not indicate the limit does not exist.
It indicates we need to perform
an algebraic technique to then hopefully determine the limit
by performing direct substitution.
Once we do this, we'll also verify the limit
by looking at a graph, as well as, a table of values.
In this case
because the numerator is the square root
of the quantity three X plus 34 minus seven.
We will rationalize the numerator
by multiplying the numerator and denominator
by the conjugate of the numerator.
So let's take it by writing the original function.
And now we will multiply the numerator and denominator
by the conjugate of the numerator,
which is the square root of the quantity
three X plus 34 plus seven.
Starting in the numerator,
we have the squared root of the quantity,
three X plus 34 times itself,
which simplifies to,
just three X plus 34.
Next, because you are multiplying conjugates,
the next two products will be opposites
and some to zero.
We have the square root of the quantity
three X plus 34 times positive seven
plus negative seven times the square root
of the quantity three X plus 34.
Those two products sum to zero.
And then we have negative seven times positive seven
which is negative 49,
giving us minus 49.
And now for the denominator,
let's go ahead and factor
X squared minus eight X plus 15.
The factor's a positive 15 that added negative eight
are negative three and negative five.
Giving us factors
of X minus three and X minus five.
And we still have times the quantity of the square root
of the quantity three X plus 34 plus seven.
And now simplify the numerator,
34 minus 49 is 15.
This simplifies to three X minus 15.
Let's go ahead and factor out the common factor of three.
Which gives us three times the quantity X minus five,
over the same denominator.
Notice in this form,
we have a common factor of X minus five
between the numerator and denominator.
By simplifying out this common factor,
we're removing the hole
in the original function at X equals five.
So now the simplified function will be continuous
at and around positive five.
And we can use this simplified function
to evaluate the original limit.
The simplified expression is now
three divided by the product of X minus three
and the square root of the quantity three X plus 34
plus seven.
Which means the original limit
is equal to the limit of the simplified expression.
As X approaches five.
And again,
because this function is continuous
at and around positive five,
we can evaluate the limit now
by performing direct substitution.
Substituting five for X,
we have three divided by the product
of five minus three
and the square root of the quantity
three times five plus 34 plus seven.
Simplifying.
We have three divided by,
well, five minus three is two.
And then under the square root
we have 15 plus 34, which is 49.
The square root of 49 is seven,
seven plus seven is 14.
The denominator simplifies to two times 14
which is equal to three twenty eighths.
So this indicates
that the original function approaches the function value
of three twenty eighths
as they approach five from the left and the right.
Let's verify this by looking at a graph,
as well as a table of values.
Looking at the graph of the original function,
most software is not going to show holes in a function.
The original function does have a hole
at X equals five,
which would be here.
But remember, this does not affect the limit.
We're only concerned about what function value
we're approaching.
As we approach five from the left
and the right.
So, we can see from the graph
as we approach five from the left and from the right
we are approaching the same function value
that is between zero and one.
Much closer to zero,
which actually is three twenty eighths.
And now look at the table of values.
As we approach five from values less than five,
we can see the decimal value we are approaching.
And if we approach X equals five
when the value is greater than five,
again, it does appear we are approaching
the same function value in decimal form.
So let's just verify that three twenty eighths
is approximately this decimal.
Notice three twenty eighths is a decimal value.
We are approaching from the left and right
of positive five, verifying our limit.
I hope you found this helpful.