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NARRATOR: Exercise 5.

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Discuss the limit of the function

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f of x equals
x squared minus 2x minus 15

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all divided by
2x squared plus 5x minus 3

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which in its factorised form
is x minus 5 times x plus 3

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all divided by 2x minus 1
times x plus 3.

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Part a.

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Discuss the limit of the function
f of x as x approaches infinity.

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So we're looking at the limit
as x approaches infinity.

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And in order to determine the limit
as x approaches infinity,

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we'll divide through

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both the numerator and denominator
of this fraction

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by the highest power of x,
so that is x squared.

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So we'll instead look at the limit
as x approaches infinity

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of one minus 2 on x minus 15
on x squared

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all divided by 2 plus 5 on x
minus 3 on x squared.

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This is easier to determine

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as looking at the limit
as x approaches infinity

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where x is in the denominator
of a fraction is easy to evaluate.

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For example, 2 on x as x approaches
infinity is going to mean that

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the denominator of that fraction
gets larger and larger and larger,

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and therefore the value of
the entire fraction approaches 0.

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So 2 on x, 15 on x squared,
5 on x and 3 on x squared

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are all going to approach 0
as x approaches infinity,

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therefore leaving one-half.

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So the limit of f of x
as x approaches infinity is one-half.

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Part b.

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Discuss the limit of the function
f of x as x approaches 5.

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So in this case, it's easiest

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to look at the function
in its factorised form.

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And we can cancel down
the factorised form

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simply to make
the calculation easier,

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although we'd get the same result
in the end regardless.

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But we can note that this function
is actually defined at x equals 5.

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So given that the function
is defined at x equals 5,

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the limit as x approaches 5
can be determined

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simply by substituting x equals 5
into this function.

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So we get 5 minus 5
divided by 10 minus 1

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which is 0 over 9, or 0.

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So the limit of f of x

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as x approaches 5 is in fact 0.

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Part b.

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Discuss the limit of the function
f of x as x approaches negative 3.

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Again, if we consider the function
in its factorised form,

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we can see that simply substituting
in x equals negative 3 at this point

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is going to be impossible,

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that is the function is undefined
at x equals negative 3.

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But we can still evaluate
the limit of this function

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by manipulating the function
a little,

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and that is noticing that

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we can cancel down
that common bracket of x plus 3

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to simply look at the limit
as x approaches negative 3

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of x minus 5 over 2x minus 1.

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So now we can substitute
x equals negative 3,

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and we determine that the limit

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of f of x as x approaches negative 3

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is eight-sevenths.

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Part d.

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Discuss the limit of the function
f of x as x approaches one-half.

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So again, we'll look
in factorised form

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and again we know
we're going to have the problem

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that the function is not actually
defined at x equals a half.

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So since f of x is not defined
at x equals a half,

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we must consider values
close to one-half.

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And we cannot simply cancel down
as we did in the previous example.

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So we look at values close to
one-half and we look at the fact that

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as x takes values close to
but greater than one-half,

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the values of f of x
are very large and positive,

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ie, we could say that
f of x approaches infinity

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as x approaches half from above.

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As x takes values close to
but less than one-half,

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the values of f of x
are very large and negative,

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that is that f of x
approaches negative infinity

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as x approaches half from below.

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So since the limit

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as x approaches half from above

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is not equal to the limit

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as x approaches half from below,

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then we say the limit as x approaches

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a half of f of x doesn't exist.

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Part e.

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Discuss the limit of the function
f of x as x approaches 0.

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So this time we can see that the
function does in fact exist at 0,

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and so calculating the limit
as x approaches 0

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is simply a case of substituting
x equals 0 into the expression.

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And we get that negative 15
divided by negative 3,

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and so the limit of f of x
as x approaches 0 is 5.