Video Transcript
Find the limit as 𝑥 approaches
negative one of 𝑥 minus six times 𝑥 squared plus two 𝑥 plus one all divided by 𝑥
squared minus six 𝑥 minus seven.
In this question, we’re asked to
evaluate the limit as 𝑥 approaches negative one of a function. And we can see the numerator of
this function is a cubic polynomial and the denominator is a quadratic
polynomial. This means this function is a
rational function; it’s the quotient between two polynomials. So as we’re evaluating the limit of
a rational function, we can attempt to do this by direct substitution. We need to substitute our value of
𝑥 is equal to negative one into our rational function. And doing this gives us negative
one minus six times negative one squared plus two times negative one plus one all
divided by negative one squared minus six times negative one minus seven.
And if we were to evaluate the
numerator and denominator of this expression separately, we would get zero divided
by zero. This is an indeterminant form. This means we can’t evaluate this
limit by using direct substitution alone. We’re going to need to use some
other form of manipulation. And since we’re dealing with
polynomials, we can try factoring our numerator and our denominator. So we want to fully factor the
quadratic in our numerator and the quadratic in our denominator.
There’s a few different ways of
doing this. We could use the quadratic formula
or we could use the quadratic solver on our calculator. However, there is another useful
method we’ll go over. Remember when we substituted 𝑥 is
equal to negative one into both of our quadratics, we got zero. And the factor theorem tells us if
negative one is a root of a polynomial, then 𝑥 plus one must be a factor of that
polynomial. Therefore, we must have 𝑥 plus one
is a factor of both of our quadratics. We can then use this to factor both
of our quadratics.
Let’s start with our
denominator. If 𝑥 plus one has to be a factor
of this quadratic, then our first term must be 𝑥 since 𝑥 times 𝑥 is equal to 𝑥
squared. And our constants must multiply
together to give us negative seven. So the other factor must be 𝑥
minus seven. And we can do exactly the same with
the quadratic in our numerator. Since the two constants multiply
together to give us one, our constant in our second factor will be one. And our two 𝑥 term should multiply
to give us 𝑥. This means our factor will be 𝑥
plus one. We can then use this to rewrite our
limit.
By fully factoring our numerator
and our denominator, we were able to rewrite the limit given to us in the question
as the limit as 𝑥 approaches negative one of 𝑥 minus six times 𝑥 plus one
multiplied by 𝑥 plus one all divided by 𝑥 minus seven times 𝑥 plus one. Now we still can’t evaluate this by
using direct substitution. If we were, we would have a factor
of zero in both our numerator and our denominator. So we would still get the
indeterminate form zero divided by zero.
But remember, when we’re looking
for the limit as 𝑥 approaches negative one of a function, we want to know what
happens when 𝑥 gets closer and closer to negative one. So we don’t need 𝑥 to be equal to
negative one. But if 𝑥 is not equal to negative
one, then 𝑥 plus one is not equal to zero. So we can cancel the shared factor
of 𝑥 plus one in our numerator and our denominator. And this won’t change the value of
our limit. So this is the same as the limit as
𝑥 approaches negative one of 𝑥 minus six times 𝑥 plus one all divided by 𝑥 minus
seven. And once again, this is the limit
of a rational function, so we can try doing this by using direct substitution. We’ll substitute 𝑥 is equal to
negative one into our rational function.
Doing this gives us negative one
minus six times negative one plus one all divided by negative one minus seven. And if we evaluate this, we see our
numerator has a factor of zero, so it’s equal to zero, and our denominator evaluates
to give us negative eight. Therefore, this equals zero divided
by negative eight, which is of course just equal to zero. Therefore, by fully factoring our
rational function, simplifying, and then using direct substitution, we were able to
show the limit as 𝑥 approaches negative one of 𝑥 minus six times 𝑥 squared plus
two 𝑥 plus one all divided by 𝑥 squared minus six 𝑥 minus seven is equal to
zero.
