Video: AP Calculus AB Exam 1 β€’ Section I β€’ Part A β€’ Question 1

Find the lim_(π‘₯ β†’ 3) (π‘₯Β² + 3π‘₯ βˆ’ 18)/(π‘₯Β² βˆ’ 9).

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Video Transcript

Find the limit as π‘₯ tends to three of π‘₯ squared plus three π‘₯ minus 18 over π‘₯ squared minus nine.

Let’s first establish why we can’t yet evaluate the expression when π‘₯ is equal to three. We’ll look at the denominator of our fraction. If we let π‘₯ be equal to three, the denominator becomes three squared minus nine. Three squared is nine and nine minus nine is zero. And we know that if we have a fraction whose denominator is zero, our solution is undefined. And so, instead, we’re going to find a way to manipulate our expression. And we do that by factoring both the numerator and the denominator of our fraction to see whether anything will simplify.

Let’s first factor π‘₯ squared plus three π‘₯ minus 18. This is a quadratic expression. And the three terms in our expression share no common factors other than one. And since the first term is π‘₯ squared, we know that this expression factors into two parentheses, in the front of which is the letter π‘₯. To find the numerical part of our parentheses, we need to find two numbers whose product is negative 18 and whose sum is three. Let’s begin by listing the factor pairs of negative 18.

Remember, a positive multiplied by a negative is a negative. So we could have negative one times 18 and one times negative 18. We could have negative two times nine and two times negative nine. And our final factor pairs are negative three and six and three and negative six. So which of these have a sum of three? Well, we can see that negative three plus six is positive three. And this means that π‘₯ squared plus three π‘₯ minus 18 factors to π‘₯ minus three multiplied by π‘₯ plus six.

And it’s always sensible to check our answers by redistributing the parentheses. We multiply the first term in each parenthesis; π‘₯ times π‘₯ is π‘₯ squared. We multiply the outer terms; π‘₯ times six is six π‘₯. The product of the inner terms is negative three π‘₯. And then we multiply the last terms; negative three times six is negative 18. Collecting like terms, and we see this simplifies to π‘₯ squared plus three π‘₯ minus 18, which is what we required.

So let’s find a way to factor π‘₯ squared minus nine. Now this can be performed in much the same way as π‘₯ squared plus three π‘₯ minus 18. However, noticing that this is a difference of two squares expression can save us some time. An expression of the form π‘Ž squared minus 𝑏 squared can be factored to π‘Ž minus 𝑏 times π‘Ž plus 𝑏. If we compare π‘₯ squared minus nine to π‘Ž squared minus 𝑏, we can see that π‘Ž must be equal to π‘₯. 𝑏 is equal to the square root of nine, which is three. And this means that we can write π‘₯ squared minus nine as π‘₯ minus three times π‘₯ plus three.

And once again, we check by redistributing the parentheses. And we get π‘₯ squared plus three π‘₯ minus three π‘₯ minus nine which is π‘₯ squared minus nine, as required. We’re now going to replace the numerator and the denominator in our limit with their factored equivalents. That gives us the limit as π‘₯ tends to three of π‘₯ minus three times π‘₯ plus six over π‘₯ minus three times π‘₯ plus three.

And notice, the numerator and the denominator of our fraction have a common factor. It’s π‘₯ minus three. So we divide through by π‘₯ minus three. And we now need to evaluate the limit as π‘₯ tends to three of π‘₯ plus six over π‘₯ plus three. We can now substitute π‘₯ is equal to three into π‘₯ plus six over π‘₯ plus three. And we get three plus six over three plus three, which is nine over six. Notice that both nine and six are multiples of three. So we can simplify this fraction, and we get three over two. So the limit as π‘₯ tends to three of π‘₯ squared plus three π‘₯ minus 18 over π‘₯ squared minus nine is three over two.

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