University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 4 - Practice Exercises - Page 277: 61



Work Step by Step

Consider $f(x)=\lim\limits_{x \to 1}\dfrac{x^2+3x-4}{x-1}=\lim\limits_{x \to 1} \dfrac{a(x)}{b(x)}$ and $a(1)=0, b(1)=0$ Then $a'(x)=2x+3; b'(x)=1$ Thus, $f(1)=\dfrac{0}{0}$ This shows an Inderminate form of the limit, so apply L-hospital's rule: $\lim\limits_{x \to l}\dfrac{a(x)}{b(x)}=\lim\limits_{x \to l}\dfrac{a'(x)}{b'(x)}$ Thus, $\lim\limits_{x \to 1}\dfrac{2x+3}{1}=5$
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