# Question Video: Finding the Limit of a Rational Function by Factoring a Sum or Difference of Two Cubes Mathematics • Higher Education

Find lim_(π₯ βΆ β2) ((π₯Β³ + 8)/(π₯ + 2)).

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### Video Transcript

Find the limit as π₯ approaches negative two of π₯ cubed plus eight all divided by π₯ plus two.

In this question, weβre asked to evaluate the limit of the quotient of two polynomials. This is a rational function. And we recall we can attempt to evaluate the limit of any rational function by using direct substitution. However, if we substitute π₯ is equal to negative two into this function, we get negative two cubed plus eight all divided by negative two plus two, which if we evaluate gives us zero divided by zero, an indeterminate form. Therefore, we canβt evaluate this limit by using direct substitution alone.

Instead, weβre going to need to use some other method. One way of evaluating this limit is to notice that negative two is a root of the polynomial in our numerator and itβs a root of the polynomial in our denominator. And by using the remainder theorem, this tells us that π₯ plus two is a factor of both the polynomial in the numerator and the denominator. And of course we already knew this for our denominator. However, we can use this to factor our numerator. And thereβs a few different ways of doing this. One way of doing this would be to use algebraic division. We would divide π₯ cubed plus eight by π₯ plus two.

However, since we know when we divide a cubic polynomial by a linear polynomial we get a quadratic polynomial, we can write this as follows. π₯ cubed plus eight will be equal to π₯ plus two multiplied by some quadratic ππ₯ squared plus ππ₯ plus π. We can then distribute our parentheses on the right-hand side of our equation. We get ππ₯ cubed plus ππ₯ squared plus ππ₯ plus two ππ₯ squared plus two ππ₯ plus two π. We can then simplify this further by taking out the shared factors of π₯ on the right-hand side of our equation. We get ππ₯ cubed plus π plus two π multiplied by π₯ squared plus π plus two π multiplied by π₯ plus two π. And remember, this expression is equal to π₯ cubed plus eight. We can find the values of π, π, and π by equating coefficients on both sides of our equation.

Letβs start by equating the coefficients of π₯ cubed. On the left-hand side of our equation, the coefficient of π₯ cubed is one. On the right-hand side of our equation, the coefficient of π₯ cubed is π. Therefore, our value of π is one. And we can substitute this into the expression on the right-hand side of our equation. Doing this then gives us the following equation.

Now, letβs equate the constant terms on both sides of our equation. We see that π is equal to two π. Therefore, π must be equal to four. And once again we can substitute this value of π back into our equation. This then gives us the following equation. Finally, we can equate the coefficients of π₯ squared on both sides of the equation or the coefficients of π₯ on both sides of the equation to find the value of π. However, on the left-hand side of our equation, thereβs no π₯ squared or π₯-term. So the coefficients of these terms are zero. And therefore, for the coefficients on the right-hand side of our equation to match, these two coefficients have to be zero. π plus two is equal to zero, and four plus two π is equal to zero, which means π is negative two.

And now we can substitute the value of π equal to one, the value of π equal to negative two, and the value of π equal to four into our factored expression. This gives us that π₯ cubed plus eight is equal to π₯ plus two multiplied by π₯ squared minus two π₯ plus four. This is exactly the same answer we wouldβve got if weβd used algebraic division.

We can now use this to help us evaluate our limit. Weβll substitute this expression for the numerator inside of our limit. Doing this, we get the limit as π₯ approaches negative two of π₯ plus two multiplied by π₯ squared minus two π₯ plus four all divided by π₯ plus two. And now we can simplify this limit. We just need to recall weβre taking the limit as π₯ approaches negative two. This means weβre interested in what happens to the output of our function as our values of π₯ get closer and closer to negative two. The output of the function when π₯ is equal to negative two will not affect its limit.

Therefore, we can cancel the shared factor of π₯ plus two in the numerator and the denominator of this expression. Because when π₯ is not equal to negative two, π₯ plus two divided by π₯ plus two is just equal to one. Therefore, weβve rewritten our limit as the limit as π₯ approaches negative two of π₯ squared minus two π₯ plus four.

This is then the limit of a polynomial. And we can evaluate the limits of polynomials by using direct substitution. Substituting π₯ is equal to negative two into our quadratic gives us negative two all squared minus two times negative two plus four, which we can evaluate is equal to 12. Therefore, we were able to show the limit as π₯ approaches negative two of π₯ cubed plus eight all divided by π₯ plus two is equal to 12.

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