Question Video: Finding the Limit of a Rational Function at a Point Using Factorisation Mathematics • Higher Education

Find lim_(π‘₯ β†’ βˆ’4)((π‘₯Β³ + 64)/(2π‘₯Β² + 6π‘₯ βˆ’ 8)).

04:52

Video Transcript

Find the limit as π‘₯ approaches negative four of π‘₯ cubed plus 64 all divided by two π‘₯ squared plus six π‘₯ minus eight.

In this question, we’re asked to evaluate the limit as π‘₯ 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 π‘₯ 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, π‘₯ cubed plus 64. Remember, when we substituted π‘₯ 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 π‘₯ 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 π‘Žπ‘₯ squared plus 𝑏π‘₯ plus 𝑐. And there’s a few different ways we could find the values of π‘Ž, 𝑏, and 𝑐. For example, we could divide π‘₯ cubed plus 64 by π‘₯ 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 π‘₯ cubed term will be π‘Žπ‘₯ cubed. This is supposed to be equal to π‘₯ cubed. Therefore, our value of π‘Ž must be equal to one. So the coefficient of π‘₯ 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 π‘₯ 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 π‘₯ squared terms, we get 𝑏π‘₯ squared plus four π‘₯ squared. However, on the left-hand side of our equation, we see we don’t have an π‘₯ squared term. In other words, the coefficient of π‘₯ squared on the left-hand side of our equation is zero.

The next thing we’ll do is find the coefficient of π‘₯ squared on the right-hand side of our equation. We’ll take out the constant factor of π‘₯ squared. This gives us zero π‘₯ 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 π‘₯ cubed plus 64 to give us π‘₯ plus four times π‘₯ squared minus four π‘₯ 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 π‘₯ squared plus six π‘₯ 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 π‘₯ squared plus three π‘₯ 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 π‘₯ plus four times π‘₯ minus one.

And now we can see both our numerator and our denominator share a factor of π‘₯ 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 π‘₯ approaches negative four of π‘₯ plus four times π‘₯ squared minus four π‘₯ plus 16 all divided by two times π‘₯ plus four multiplied by π‘₯ minus one. And the reason this is useful is we’re evaluating the limit as π‘₯ approaches negative four. We want to know what happens to our function as π‘₯ gets closer and closer to negative four.

But if we’re only interested in what’s happening when π‘₯ is getting close to negative four, then π‘₯ is not equal to negative four. So we can cancel the shared factor of π‘₯ 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 π‘₯ approaches negative four of π‘₯ squared minus four π‘₯ plus 16 all divided by two times π‘₯ 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 π‘₯ 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 π‘₯ approaches negative four of π‘₯ cubed plus 64 all divided by two π‘₯ squared plus six π‘₯ minus eight is equal to negative 24 divided by five.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.