# Video: Finding the Limit of a Rational Function

Find lim_(π₯ β 3) (π₯β΄ β 81)/(π₯Β³ β 27).

05:59

### Video Transcript

Find the limit as π₯ approaches three of π₯ to the fourth power minus 81 divided by π₯ cubed minus 27.

We see the question is asking us to calculate the limit as π₯ approaches three of a quotient of polynomials. We call these rational functions. And we know we can attempt to evaluate rational functions by using direct substitution. Substituting π₯ is equal to three, we get three to the fourth power minus 81 divided by three cubed minus 27. And if we evaluate this expression, we see we get zero divided by zero. This is an indeterminate form. This tells us that we canβt evaluate our limit directly by using substitution. So weβre going to need to try a different method to evaluate this limit.

Thereβs a few different ways of evaluating this limit. The easiest way is to notice that our rational function is equal to π₯ to the fourth power minus three to the fourth power divided by π₯ cubed minus three cubed. And thereβs actually a standard way to evaluate limits of this form. We can actually evaluate the limit as π₯ approaches π of π₯ to the πth power minus π to the πth power divided by π₯ to the πth power minus π to the πth power as π divided by π times π to the power of π minus π. And we can see weβve rewritten our rational function in this form, where π is equal to four, π is equal to three, and π is equal to three.

So using this result with these values of π, π, and π, we get that our limit is equal to four-thirds times three to the power of four minus three. And we can simplify this. Three to the power of four minus three is three to the first power, which is just equal to three. And then four-thirds times three is just equal to four. So thatβs the easiest way of showing that the limit given to us in the question is equal to four.

We can also do this by using the factor theorem. When we used direct substitution to attempt to evaluate this limit, we saw that when we substituted π₯ is equal to three into the polynomial in our numerator and the polynomial in our denominator, we saw that we got zero for both of these. And the factor theorem tells us if we substitute π₯ is equal to three into our polynomial and this evaluates to give us zero, then π₯ minus three must be a factor of this polynomial.

So we can attempt to evaluate this limit by taking out our factors of π₯ minus three. So we want to take out a factor of π₯ minus three from the polynomial in our numerator and the polynomial in our denominator. Thereβs a few different ways of doing this. For example, we could use algebraic division. However, thereβs a simpler method for our numerator.

We see that our numerator is in fact a difference between squares. We see π₯ to the fourth power is π₯ squared squared and 81 is nine squared. And we know we can factor the difference between squares, π squared minus π squared, as π minus π multiplied by π plus π. So we can factor our numerator as π₯ squared minus nine times π₯ squared plus nine. And then we can see again that π₯ squared minus nine is a difference between squares, since nine is equal to three squared. So we can factor this by using difference between squares again. We get π₯ squared minus nine is equal to π₯ minus three times π₯ plus three.

So weβve now factored our numerator to show π₯ to the fourth power minus 81 is equal to π₯ minus three times π₯ plus three times π₯ squared plus nine. We now want to take our factor of π₯ minus three from our denominator. Thereβs a few different ways of doing this. Weβre going to notice that we need a linear factor multiplied by a quadratic factor to give us our cubic. But we donβt know the coefficients of this quadratic. Weβll call them π, π, and π.

To find the values of π, π, and π, weβre going to multiply our linear term and our quadratic term and then equate coefficients with our cubic term. Letβs start by equating the constant. We see the constant not cubic is negative 27. And when we multiply our linear factor and our quadratic factor, we see the constant will be negative three times π. So we want to equate these two constants. We have negative 27 is equal to negative three π. And we divide through by negative three to see that π is equal to nine.

Letβs do the same by equating the coefficients of π₯ cubed. In our cubic, the coefficient of π₯ cubed is one. And when we multiply our linear factor and our quadratic factor, we see the coefficient of π₯ cubed will be π times one, which is just π. So again, we equate these coefficients, and we see that π is equal to one.

We now need to find the value of π. And it might seem tricky to see how to do this, since we appear to have run out of coefficients to compare. However, we remember that we can just add the term zero π₯ squared. This doesnβt change the value of our cubic.

And now we can compare the coefficients of π₯ squared. In our cubic, the coefficient of π₯ squared is zero. And we must be careful. When we multiply our linear term and our quadratic term, thereβs two ways of getting coefficients of π₯ squared. When we multiply these two factors, our π₯ squared term will be ππ₯ squared minus three π₯ squared. So the coefficient of π₯ squared will be π minus three.

Just as we did before, weβll equate these coefficients. We get zero is equal to π minus three. And we can rearrange just to see that π is equal to three. So weβve now factored our denominator. This gives us π₯ cubed minus 27 is equal to π₯ minus three times π₯ squared plus three π₯ plus nine.

To evaluate this limit, weβre just going to cancel the shared factor of π₯ minus three in our numerator and our denominator. We can do this since it wonβt change the limit as π₯ approaches three of this function. So weβre now trying to evaluate the limit as π₯ approaches three of π₯ plus three times π₯ squared plus nine divided by π₯ squared plus three π₯ plus nine.

This is a rational function. So we can attempt to evaluate this by direct substitution. Substituting π₯ is equal to three, we get three plus three times three squared plus nine divided by three squared plus three times three plus nine. And if we evaluate this expression, we get four.

So weβve shown in two different ways the limit as π₯ approaches three of π₯ to the fourth power minus 81 divided by π₯ cubed minus 27 is equal to four.