University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 4 - Section 4.5 - Indeterminate Forms and L'Hôpital's Rule - Exercises - Page 248: 33



Work Step by Step

L'Hospital's rule states that $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{A'(x)}{B'(x)}$ Here, $\lim\limits_{x \to 0^{+}} f(0)=\dfrac{0}{0}$ This shows an indeterminate form of a limit, so we need to use L'Hospital's rule. $ \lim\limits_{x \to 0^{+}} \dfrac{2x+2/x^2+2x}{1/x}=\lim\limits_{x \to 0^{+}} \dfrac{2x^2+2x}{x^2+2x}=\dfrac{0}{0}$ Again apply L'Hospital's rule. $\lim\limits_{x \to 0^{+}} \dfrac{4x+2}{2x+2}=\dfrac{2}{2}=1$
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