Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 2 - Limits - 2.3 Techniques for Computing Limits - 2.3 Exercises - Page 78: 73

Answer

$5a^4$

Work Step by Step

We have to determine $L=\lim\limits_{x \to a} \dfrac{x^5-a^5}{x-a}$ Use the factorization formula: $x^n-a^n=(x-a)(x^{n-1}+x^{n-2}a+......+xa^{n-2}+a^{n-1})$ for $n=5$: $L=\lim\limits_{x \to a} \dfrac{(x-a)(x^4+x^3a+x^2a^2+xa^3+a^4)}{x-a}$ Simplify: $L=\lim\limits_{x \to a} (x^4+x^3a+x^2a^2+xa^3+a^4)$ Determine the limit: $L=a^4+a^3(a)+a^2(a^2)+a(a^3)+a^4=a^4+a^4+a^4+a^4+a^4=5a^4$
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