We're asked to evaluate the limit.
Notice how we have the limit of a complex fraction
as t approaches three.
If we try to perform direct substitutions
to determine the limit,
notice how in the numerator we have 1/5 minus 1/5,
which is zero.
In the denominator, we have three minus three,
which is also zero,
giving us the indeterminate form of zero divided by zero.
This does not mean the limit does not exist.
We can often perform algebraic techniques
to change the form of the function
that will then allow us to determine the limit
by performing direct substitution.
In this case, we'll be using method three
where we'll use the least common denominator.
We can approach this two ways.
One way is just to subtract the fractions in the numerator
as the first step.
A quicker method would be to notice
that the least common denominator is
five times the quantity t plus two
and then multiply the top and bottom of the complex fraction
by that product.
I'm going to go ahead and show the longer steps
of subtracting the fractions in the numerator.
So beginning with the complex fraction,
if we needed to subtract the fractures in the numerator,
we would have a common denominator
or least common denominator
of five times the quantity t plus two.
And therefore, we would multiply the numerator
and denominator of the first fraction by five.
We would multiply the numerator
and denominator of the second fraction by t plus two.
And now let's go ahead and subtract.
Again, we have a common denominator
of five times the quantity t plus two.
The numerator is now going to be five
minus the quantity t plus two.
Let's go ahead and simplify five
minus the quantity t plus two.
We would have five minus t minus two.
And five minus two is three.
Let's write that as negative t plus three.
From here remember a fraction bar represents division,
dividing by t minus three is equivalent
to multiplying by the reciprocal
of one divided by the quantity t minus three.
Thus, write the complex fraction as negative t plus three
divided by five times the quantity t plus two
times one over t minus three.
Now from here, if we analyze negative t plus three
and t minus three, these are opposites.
Notice for negative t plus three, the t is negative.
The three is positive.
And for t minus three,
the t is positive and the three is negative.
So we'll factor out negative one from negative t plus three,
which will give us negative one
times the quantity t minus three.
Let's rewrite this one more time.
We have negative one times the quantity t minus three
all over the product of five and t plus two
times one over t minus three.
From here, we can see we have a common factor
of t minus three between the numerator and denominator.
T minus three divided by itself simplifies to one.
So the complex fraction simplifies to negative one
divided by the product of five and t plus two,
which means you can write the original limit
as the limit of negative one
divided by the product of five and t plus two
as t approaches three.
Notice the simplified function
is continuous around t equals three
and therefore, we can determine the limit
by performing direct substitution.
Before we do this though, I do want to show
how we could have simplified this complex fraction
in a few less steps.
So beginning with the original complex fraction,
if we recognize the least common denominator
is five times the quantity t plus two,
it's faster if we multiply
the numerator of the complex fraction
by five times the quantity t plus two
and multiply the denominator
by the same product of five and t plus two.
Let's go ahead and do this.
One over t plus two times five times t plus two
is just five.
And then we have minus 1/5
times five times t plus two is t plus two.
In the denominator, we have five times t minus three
times t plus two.
Simplifying in the numerator here,
we know that five minus the quantity t plus two
is going to give us negative t plus three.
If we factor out the negative one,
we are going to have negative one
times the quantity t minus three, just like we had above.
So we would have negative one
times the quantity t minus three.
In the denominator, we still have five
times the quantity t minus three
times the quantity t plus two.
And just as above, t minus three divided by itself
simplifies to one leaving us with the same result
of negative one over the product of five and t plus two.
And again, the simplified function is continuous
around t equals three
and therefore, we can perform direct substitution
to find the limit.
We have negative one over the product of five.
And if t is three, we have three plus two, or five,
which gives us negative 1/25 as the limit.
Let's see if we can verify this
using a table of values as well as a graph.
Since we have the limit as t approaches three,
we have to approach the same function value
as we approach t equals three from the left and the right.
So approaching three from the left
are values less than three.
We can see the function values
are approaching negative 0.04, which is negative 1/25.
And if we approach three from values greater than three,
as we see here, we are approaching the same function value
of negative 0.04 or negative 1/25.
Or looking at the graph,
if we draw a vertical line at t equals three,
it's going to be hard to tell
but as we approach three from the left and from the right,
we are approaching a value that's just less than zero.
We are approaching the function value of negative 1/25.
I hope you found this helpful.