Video: Finding the Limit of a Rational Function Using Factorization of the Difference of Two Squares

Find lim_(π‘₯β†’2) ((π‘₯Β² βˆ’ 4)/(π‘₯Β² βˆ’ 9) Γ— (π‘₯ βˆ’ 3)/(π‘₯ βˆ’ 2)).

03:08

Video Transcript

Find the limit as π‘₯ tends to two of the expression π‘₯ squared minus four over π‘₯ squared minus nine multiplied by π‘₯ minus three over π‘₯ minus two.

Now, we might be tempted to answer this question by substituting π‘₯ equals two into the expression. Let’s see what happens if we go ahead and do that. Our expression becomes two squared minus four over two squared minus nine multiplied by two minus three over two minus two. Now of course, two squared is equal to four. So the first fraction becomes four minus four, which is zero, over four minus nine, which is negative five. And there’s no problems with this so far.

The second fraction however becomes negative one over zero. We know that we can’t calculate negative one divided by zero. It’s undefined. So we’re going to need to find an alternative method for solving this problem. Instead, we’re going to look to simplify our expression by factorising where possible.

We’ll begin by factorising the expression π‘₯ squared minus two [four]. This is a special example of the difference of two squares. We have two square numbers separated by a subtraction symbol. We therefore find the square root of each part. The square root of π‘₯ squared is π‘₯. So we add an π‘₯ at the front of each bracket. The square root of four is two. So we added two as the numerical part of each bracket. And with the difference of two squares, we make one of these plus two and one of these negative two. And we’ll see why when we check our answer.

We can check our answer by multiplying our brackets back out again. We multiply the first term in each bracket. π‘₯ multiplied by π‘₯ is π‘₯ squared. We multiply the outer terms. π‘₯ multiplied by negative two is negative two π‘₯. Then, the inner terms, two multiplied by π‘₯ is two π‘₯. And the last terms, two multiplied by negative two is negative four. And we can see now that negative two π‘₯ plus two π‘₯ is zero. So the expansion of π‘₯ plus two by π‘₯ minus two does indeed give us π‘₯ squared minus four as required.

We’ll repeat this process for π‘₯ squared minus nine. Once again, we have an π‘₯ at the front of each bracket. And the square root of nine is three. So this factorizes to π‘₯ plus three multiplied by π‘₯ minus three. And so now, we’re finding the limit as π‘₯ tends to two of π‘₯ plus two multiplied by π‘₯ minus two over π‘₯ plus three multiplied by π‘₯ minus three multiplied by π‘₯ minus three over π‘₯ minus two.

And now, we can simplify. We divide through by π‘₯ minus two. We then divide through by π‘₯ minus three. And we can see that we’re now multiplying π‘₯ plus two over π‘₯ plus three by one. And we have the limit as π‘₯ tends to two of π‘₯ plus two over π‘₯ plus three. And now, we can evaluate this. Let’s substitute two into our expression. We get two plus two over two plus three, which is four over five.

And we see that the limit as π‘₯ tends to two of π‘₯ squared minus four over π‘₯ squared minus nine multiplied by π‘₯ minus three over π‘₯ minus two is four-fifths.

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