Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 1 - Limits and Their Properties - 1.5 Exercises: 43



Work Step by Step

$\lim\limits_{x\to0^+}\dfrac{2}{\sin{x}}=\dfrac{2}{\sin{0^+}}=\dfrac{2}{0^+}=+\infty.$ Note: for $\theta$ (in radians) close to $0$ , $\sin{\theta}\approx\theta.$ Since, $0^+$ is extremely close to $0$, then $\sin{0^+}=0^+$
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