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: 72

Answer

7

Work Step by Step

We have to determine $L=\lim\limits_{x \to -1} \dfrac{x^7+1}{x+1}$ Use the factorization formula: $x^n-a^n=(x-a)(x^{n-1}+x^{n-2}a+......+xa^{n-2}+a^{n-1})$ for $n=7$ and $a=-1$: $L=\lim\limits_{x \to -1} \dfrac{x^7-(-1)^7}{x+1}=\lim\limits_{x \to -1} \dfrac{(x+1)(x^6-x^5+x^4-x^3+x^2-x+1)}{x+1}$ Simplify: $L=\lim\limits_{x \to -1} (x^6-x^5+x^4-x^3+x^2-x+1)$ Determine the limit: $L=(-1)^6-(-1)^5+(-1)^4-(-1)^3+(-1)^2-(-1)+1=1+1+1+1+1+1+1=7$
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