Question Video: Finding the Limit of Rational Functions at a Point Mathematics • Higher Education

Find lim_(𝑥 → −1) ((𝑥 − 6)(𝑥² + 2𝑥 + 1))/(𝑥² − 6𝑥 − 7).

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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.

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