Question Video: Finding the Limit of a Rational Function | Nagwa Question Video: Finding the Limit of a Rational Function | Nagwa

# Question Video: Finding the Limit of a Rational Function Mathematics

Find lim_(π₯ βΆ 1) [((π₯β΄ β 1)β΄/(π₯Β³ β 1)Β³) Γ (1/(π₯βΆ β 1))].

04:46

### Video Transcript

Find the limit as π₯ approaches one of π₯ to the fourth power minus one all raised to the fourth power all divided by π₯ cubed minus one all cubed multiplied by one over π₯ to the sixth power minus one.

In this question, weβre asked to evaluate the limit of a rational function. This is because the numerator is a polynomial and the denominator is a polynomial. So, we could attempt to do this by using direct substitution. However, if we substitute π₯ is equal to one into our limit, in our numerator, we get a factor of one to the fourth power minus one, which is zero. And in our denominator, we get a factor of one cubed minus one, which is also zero. So, this evaluates to give us the indeterminate form zero divided by zero.

So, we canβt just evaluate this limit by direct substitution alone. Weβre going to need to manipulate our limit. One way of doing this would be to look for factors of π₯ minus one in the numerator and denominator. We could then cancel these shared factors, and this would help us evaluate this limit. However, thereβs actually an easier method if we can rewrite this limit in terms of the limit of a difference of powers. And to do this, we recall the following limit result. For any real constants π, π, and π, where π is not equal to zero, the limit as π₯ approaches π of π₯ to the πth power minus π to the πth power all divided by π₯ to the power of π minus π to the power of π is equal to π over π multiplied by π to the power of π minus π. And this holds true provided π to the πth power, π to the πth power, and π to the power of π minus π all exist.

We want to rewrite our limit in terms of this limit result. First, we note weβre taking the limit as π₯ approaches one, so our value of π is going to be equal to one. Next, we also note that one raised to any power is equal to one. So, in each of these expressions, we can rewrite one to match the exponent of π₯. For example, in the numerator of the fraction on the left, we can rewrite one as one to the fourth power. This is to match the numerator of our limit result; we need π to the power of π. Similarly, in our denominator, we can rewrite one as one cubed. This is to match the denominator of our limit result. Doing this can help us notice instances where we can apply this limit result to simplify our limit. For example, we can see the limit result in our first factor. The only problem is our numerator is raised to the fourth power where our denominator is cubed.

To apply this limit result, weβre going to want to match the exponents so we can take the power out. To do this, we need to take one factor of π₯ to the fourth power minus one and put it in our second factor. This gives us the limit as π₯ approaches one of π₯ to the fourth power minus one to the fourth power all cubed all over π₯ cubed minus one cubed all cubed multiplied by π₯ to the fourth power minus one to the fourth power all over π₯ to the sixth power minus one to the sixth power.

Now, the second factor inside our limit is exactly in the form of our limit result. However, the first factor is not quite in this form because weβre cubing both the numerator and denominator. So, weβll need to apply our laws of exponents. This will instead allow us to cube the entire factor. Now, our first factor is the cube of this limit result. So, we can apply the product rule for limits and the power rule for limits to evaluate our limit by using this limit result. Weβll start by applying the product rule for limits, which tells us the limit of a product of two functions is equal to the product of the limits of those two functions, provided both limits exist.

Next, weβll also apply the power rule for limits. This allows us to take the cube outside of our limit. And itβs worth pointing out we know the limits of both of these two functions exist, and we know how to evaluate them because we can just apply our limit result. Weβre now ready to evaluate both of these limits separately. Letβs start with our first limit. We can see our exponent π is four, our exponent π is three, and our value of π is one. Then, we just substitute these values into our limit result. We get four over three multiplied by one to the power of four minus three. And remember, we need to cube the value of this limit.

We can do exactly the same with our second limit. Our value of π is four, our value of π is six, and our value of π is one. We can then evaluate this limit by substituting these values into our limit result. We get four over six multiplied by one to the power of four minus six. Then, because both of these limits exist, the product of these two values is going to be the value of our original limit.

Now, all we need to do is evaluate this expression. First, one raised to any power is just equal to one, so we can simplify this. We get four-thirds all cubed multiplied by four over six. We can then evaluate this expression. And if we do, we get 128 divided by 81, which is our final answer.

Therefore, we were able to show the limit as π₯ approaches one of π₯ to the fourth power minus one all raised to the fourth power all divided by π₯ cubed minus one all cubed multiplied by one divided by π₯ to the sixth power minus one is equal to 128 divided by 81.