Video Transcript
Find the limit as 𝑥 approaches
negative four of 𝑥 cubed plus 64 all divided by two 𝑥 squared plus six 𝑥 minus
eight.
In this question, we’re asked to
evaluate the limit as 𝑥 approaches negative four of the quotient of two
polynomials. We call these rational
functions. And we know we can always attempt
to evaluate the limits of a rational function by using direct substitution. So we’ll try and evaluate our limit
by direct substitution. We’ll substitute 𝑥 is equal to
negative four into our rational function. This gives us negative four cubed
plus 64 all divided by two times negative four squared plus six times negative four
minus eight.
And if we calculate both the
expression in our numerator and denominator, we get zero divided by zero. We call this an indeterminate
form. So because we got an indeterminate
form, we’re going to need to try another method to evaluate this limit. And since we’re evaluating the
limit of the quotient of two polynomials, we’ll try factoring both of our
polynomials. Let’s start with the cubic
polynomial in our numerator, 𝑥 cubed plus 64. Remember, when we substituted 𝑥 is
equal to negative four into this expression, we saw that this was a root of our
polynomial. It was equal to zero.
And remember, by using the factor
theorem, if negative four is a root of our polynomial, then 𝑥 plus four is a factor
of our polynomial. And we know to get a cubic
polynomial, we need to multiply a linear polynomial by a quadratic. So we’ll call our quadratic factor
𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐. And there’s a few different ways we
could find the values of 𝑎, 𝑏, and 𝑐. For example, we could divide 𝑥
cubed plus 64 by 𝑥 plus four by using algebraic division. However, another method we could
use is to multiply the right-hand side of our equation out and then equate
coefficients.
For example, when we distribute on
the right-hand side of our equation, we can see the only 𝑥 cubed term will be 𝑎𝑥
cubed. This is supposed to be equal to 𝑥
cubed. Therefore, our value of 𝑎 must be
equal to one. So the coefficient of 𝑥 squared is
one. We can do something very similar to
find the value of 𝑐. We can see that the only constant
term will be four 𝑐. Then if we equate the constant
terms on both sides of the equation, we get that 64 must be equal to four 𝑐. Dividing through by four gives us
that 𝑐 is equal to 16.
So the constant term in our
quadratic is 16. We now need to find the value of
𝑏. This is slightly more
difficult. We’ll do this by equating the terms
with coefficients of 𝑥 squared. We can see there’ll be two on the
right-hand side of our equation. Distributing on the right-hand side
of our equation and only writing down the 𝑥 squared terms, we get 𝑏𝑥 squared plus
four 𝑥 squared. However, on the left-hand side of
our equation, we see we don’t have an 𝑥 squared term. In other words, the coefficient of
𝑥 squared on the left-hand side of our equation is zero.
The next thing we’ll do is find the
coefficient of 𝑥 squared on the right-hand side of our equation. We’ll take out the constant factor
of 𝑥 squared. This gives us zero 𝑥 squared is
equal to 𝑏 plus four. In other words, 𝑏 plus four is
equal to zero. And of course, that means the value
of the constant 𝑏 is negative four. So we’ve factored 𝑥 cubed plus 64
to give us 𝑥 plus four times 𝑥 squared minus four 𝑥 plus 16. But this won’t be enough to
evaluate our limit by using direct substitution. So we’ll also factor the polynomial
in our denominator.
We want to factor two 𝑥 squared
plus six 𝑥 plus eight. The first thing we need to notice
is each of our terms shares a factor of two. So we’ll rewrite this quadratic as
two times 𝑥 squared plus three 𝑥 minus four. Now, we want to factor this
quadratic. We could do this by using the
quadratic formula or a quadratic solver or we can use the factor theorem. Or we can notice that four
multiplied by negative one is equal to negative four and four plus negative one is
equal to three. So it factors to give us 𝑥 plus
four times 𝑥 minus one.
And now we can see both our
numerator and our denominator share a factor of 𝑥 plus four. We can use this to evaluate our
limit. First, by factoring our numerator
and our denominator, we were able to rewrite our limit as the limit as 𝑥 approaches
negative four of 𝑥 plus four times 𝑥 squared minus four 𝑥 plus 16 all divided by
two times 𝑥 plus four multiplied by 𝑥 minus one. And the reason this is useful is
we’re evaluating the limit as 𝑥 approaches negative four. We want to know what happens to our
function as 𝑥 gets closer and closer to negative four.
But if we’re only interested in
what’s happening when 𝑥 is getting close to negative four, then 𝑥 is not equal to
negative four. So we can cancel the shared factor
of 𝑥 plus four in our numerator and denominator. This tells us that the limit given
to us in the question is equal to the limit as 𝑥 approaches negative four of 𝑥
squared minus four 𝑥 plus 16 all divided by two times 𝑥 minus one. And this is the limit of a rational
function, so we can attempt to evaluate this by using direct substitution.
So we substitute in 𝑥 is equal to
negative four. This gives us negative four squared
minus four times negative four plus 16 all divided by two times negative four minus
one. And if we evaluate this expression
and simplify, we get negative 24 divided by five. Therefore, we were able to show the
limit as 𝑥 approaches negative four of 𝑥 cubed plus 64 all divided by two 𝑥
squared plus six 𝑥 minus eight is equal to negative 24 divided by five.
