Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Section 7.5 - Indeterminate Forms and L'Hopital's Rule - Exercises 7.5 - Page 410: 65

Answer

$1$

Work Step by Step

$\lim\limits_{x \to 0^{+}} f(x)=\lim\limits_{x \to 0^{+}} \dfrac{x}{\cot (\frac{\pi}{2}-x)}$ But $\lim\limits_{x \to 0^{+}} \dfrac{x}{\cot (\frac{\pi}{2}-x)}=\dfrac{0}{0}$ This shows an indeterminate form of limit, thus we will apply L-Hospital's rule such as: $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{p'(x)}{q'(x)}$ $\lim\limits_{x \to 0^{+}} \dfrac{1}{\csc^2(\frac{\pi}{2}-x)}=\lim\limits_{x \to 0^{+}} [\sin^2(\frac{\pi}{2}-x)]$ and $\sin^2(\dfrac{\pi}{2}-0)=1$
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