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: 76


a) The limit provided in part (a) is not correct. b) The limit provided in part (b) is correct.

Work Step by Step

a) L'Hospital's rule states that $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{A'(x)}{B'(x)}$ Here, $\lim\limits_{x \to 0} \dfrac{x^2-2x}{x^2-\sin x}=\lim\limits_{x \to 0} \dfrac{2x-2}{2x-\cos x}=1$ Also, $\lim\limits_{x \to 0} \dfrac{x^2-2x}{x^2-\sin x}=\lim\limits_{x \to 0} \dfrac{2x-2}{2x-\cos x} \ne \lim\limits_{x \to 0} \dfrac{2}{2+\sin x}=1$ (L'Hospital's rule can not be used because we do not have indeterminate form.) b) $\lim\limits_{x \to 0} \dfrac{x^2-2x}{x^2-\sin x}=\lim\limits_{x \to 0} \dfrac{2x-2}{2x-\cos x}=\dfrac{2(0)-2}{2(0)-\cos 0}=2$ Hence, a) The limit provided in part(a) is not correct. b) The limit provided in part(b) is correct.
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