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: 65



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, $\lim\limits_{x \to 0^{+}} f(x)=\lim\limits_{x \to 0^{+}} \dfrac{x}{\cot (\frac{\pi}{2}-x)}$ Now, $\lim\limits_{x \to 0^{+}} \dfrac{x}{\cot (\frac{\pi}{2}-x)}=\dfrac{0}{0}$ This shows an indeterminate form of the limit, so we need to use L'Hospital's rule. $\lim\limits_{x \to 0^{+}} \dfrac{1}{\csc^2(\frac{\pi}{2}-x)}=\lim\limits_{x \to 0^{+}} \sin^2(\frac{\pi}{2}-x)$ Thus, $\sin^2(\frac{\pi}{2}-0)=1$
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