NARRATOR: Exercise 5.
Discuss the limit of the function
f of x equals x squared minus 2x minus 15
all divided by 2x squared plus 5x minus 3
which in its factorised form is x minus 5 times x plus 3
all divided by 2x minus 1 times x plus 3.
Part a.
Discuss the limit of the function f of x as x approaches infinity.
So we're looking at the limit as x approaches infinity.
And in order to determine the limit as x approaches infinity,
we'll divide through
both the numerator and denominator of this fraction
by the highest power of x, so that is x squared.
So we'll instead look at the limit as x approaches infinity
of one minus 2 on x minus 15 on x squared
all divided by 2 plus 5 on x minus 3 on x squared.
This is easier to determine
as looking at the limit as x approaches infinity
where x is in the denominator of a fraction is easy to evaluate.
For example, 2 on x as x approaches infinity is going to mean that
the denominator of that fraction gets larger and larger and larger,
and therefore the value of the entire fraction approaches 0.
So 2 on x, 15 on x squared, 5 on x and 3 on x squared
are all going to approach 0 as x approaches infinity,
therefore leaving one-half.
So the limit of f of x as x approaches infinity is one-half.
Part b.
Discuss the limit of the function f of x as x approaches 5.
So in this case, it's easiest
to look at the function in its factorised form.
And we can cancel down the factorised form
simply to make the calculation easier,
although we'd get the same result in the end regardless.
But we can note that this function is actually defined at x equals 5.
So given that the function is defined at x equals 5,
the limit as x approaches 5 can be determined
simply by substituting x equals 5 into this function.
So we get 5 minus 5 divided by 10 minus 1
which is 0 over 9, or 0.
So the limit of f of x
as x approaches 5 is in fact 0.
Part b.
Discuss the limit of the function f of x as x approaches negative 3.
Again, if we consider the function in its factorised form,
we can see that simply substituting in x equals negative 3 at this point
is going to be impossible,
that is the function is undefined at x equals negative 3.
But we can still evaluate the limit of this function
by manipulating the function a little,
and that is noticing that
we can cancel down that common bracket of x plus 3
to simply look at the limit as x approaches negative 3
of x minus 5 over 2x minus 1.
So now we can substitute x equals negative 3,
and we determine that the limit
of f of x as x approaches negative 3
is eight-sevenths.
Part d.
Discuss the limit of the function f of x as x approaches one-half.
So again, we'll look in factorised form
and again we know we're going to have the problem
that the function is not actually defined at x equals a half.
So since f of x is not defined at x equals a half,
we must consider values close to one-half.
And we cannot simply cancel down as we did in the previous example.
So we look at values close to one-half and we look at the fact that
as x takes values close to but greater than one-half,
the values of f of x are very large and positive,
ie, we could say that f of x approaches infinity
as x approaches half from above.
As x takes values close to but less than one-half,
the values of f of x are very large and negative,
that is that f of x approaches negative infinity
as x approaches half from below.
So since the limit
as x approaches half from above
is not equal to the limit
as x approaches half from below,
then we say the limit as x approaches
a half of f of x doesn't exist.
Part e.
Discuss the limit of the function f of x as x approaches 0.
So this time we can see that the function does in fact exist at 0,
and so calculating the limit as x approaches 0
is simply a case of substituting x equals 0 into the expression.
And we get that negative 15 divided by negative 3,
and so the limit of f of x as x approaches 0 is 5.