Explanation Given information: lim x → − 2 3 x 2 + a x + a + 3 x 2 + x − 2 Calculation: For the limit lim x → − 2 3 x 2 + a x + a + 3 x 2 + x − 2 To exist, it is necessary that the polynomial in the numerator evaluated at x = - 2 is 0 because the denominator is zero at x = - 2. From the necessary condition we get 3 ( − 2 ) 2 + a ( − 2 ) + a + 3 = 0 12 − 2 a + a + 3 = 0 15 − a = 0 a = 15 Now let us factorize the numerator for this value of a . 3 x 2 + 15 x + 15 + 3 = 3 x 2 + 15 x + 18 = 3 x 2 + 9 x + 6 x + 18 = 3 x ( x + 3 ) + 6 ( x + 3 ) &

Bundle: Biocalculus: Calculus, Probability, And Statistics For The Life Sciences + Webassign Printed Access Card For Stewart/day's Biocalculus: Calculus For Life Sciences, 1st Edition, Multi-term

1st Edition

ISBN: 9781305779037

Author: James Stewart, Troy Day

Publisher: Cengage Learning

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Limits

When it comes to calculus, limits are considered to be a very important topic of discussion. The main importance of limit lies in expressing the value of a function as it gets closer to a certain value. Also, limits can help you to understand other topic…

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Chapter 2.4, Problem 51E

To determine

To find: a number a such that limx→−23x2+ax+a+3x2+x−2 exists and find the value of a and the value of the limit.

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Chapter 2.4, Problem 50E

Chapter 2.4, Problem 52E

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Chapter 2 Solutions

Bundle: Biocalculus: Calculus, Probability, And Statistics For The Life Sciences + Webassign Printed Access Card For Stewart/day's Biocalculus: Calculus For Life Sciences, 1st Edition, Multi-term

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a number a such that lim x → − 2 3 x 2 + a x + a + 3 x 2 + x − 2 exists and find the value of a and the value of the limit.

Solution Summary: The author explains that to find the value of a and the limit, it is necessary that the polynomial in the numerator evaluated at x = - 2 is 0 because the denominator is zero

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To find: a number a such that limx→−23x2+ax+a+3x2+x−2 exists and find the value of a and the value of the limit.

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Chapter 2 Solutions