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2) ((π₯ β 4)Β³ + 8)/(π₯ β 2).

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### Video Transcript

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

In this question, weβre asked to evaluate the limit of a function. We can see in our numerator we have a polynomial and in our denominator we have a polynomial. So this is a rational function. And we could always try to evaluate the limit of a rational function by direct substitution. Substituting π₯ is equal to two into the function, we get two minus four all cubed plus eight all divided by two minus two, which, if we evaluate, we see is zero divided by zero, which is an indeterminate form. Since this gives an indeterminate form, we canβt evaluate this limit by using direct substitution; weβre going to need to use a different method.

We need to notice the limit given to us in the question is very similar to one of our limit results. That is, the limit as π₯ approaches zero of π₯ plus π all raised to the πth power minus π to the πth power all divided by π₯ is equal to π times π to the power of π minus one. And thatβs provided π to the πth power and π to the power of π minus one both exist. But this limit result has π₯ approaching zero, and the limit weβre asked to evaluate has π₯ approaching two. So weβre going to use the substitution π¦ is equal to π₯ minus two. Then, as our values of π₯ approach two, π₯ minus two is going to be approaching zero. So our values of π¦ approach zero.

In our denominator, we have π₯ minus two, which is just going to be equal to π¦. However, in our numerator, we have π₯ minus four. So we need to find an expression for π₯ minus four. And we can find this by subtracting two from both sides of our equation for π¦. We get π¦ minus two is equal to π₯ minus four. Therefore, by using the substitution π¦ is equal to π₯ minus two, we were able to rewrite our limit as the limit as π¦ approaches zero of π¦ minus two all cubed plus eight all divided by π¦.

And this is now almost exactly in the form of our limit result. We can write it in the exact form of our limit result by noting that π¦ plus negative two is the same as π¦ minus two and eight is the same as negative one times negative two all cubed. So our value of π is negative two and our value of π is three. Therefore, our limit result tells us that this limit is equal to π multiplied by π to the power of π minus one. Substituting π is equal to negative two and π is equal to three, we get three multiplied by negative two to the power of three minus one, which we can evaluate is equal to 12. Therefore, we were able to show the limit as π₯ approaches two of π₯ minus four all cubed plus eight all divided by π₯ minus two is equal to 12.