Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 2 - Limits - 2.6 Trigonometric Limits - Exercises - Page 77: 55


The limit does not exist.

Work Step by Step

\begin{align*} \lim _{t \rightarrow 0^+}\frac{\sqrt{1-\cos t}}{t}&= \lim _{t \rightarrow 0^+}\left(\frac{1-\cos t}{t^2}\right)^{1/2}\\ &=\left( \lim _{t \rightarrow 0^+}\frac{1-\cos t}{t^2}\right)^{1/2}\\ &=\left(\frac{1}{2} \right)^{1/2}\\ &=\frac{1}{\sqrt{2}}. \end{align*} Where we used the fact that $\lim _{x \rightarrow 0^+} \frac{1-\cos x}{x^2}=\frac{1}{2}$. By the same way, we have \begin{align*} \lim _{t \rightarrow 0^-}\frac{\sqrt{1-\cos t}}{t}&=- \lim _{t \rightarrow 0^-}\left(\frac{1-\cos t}{t^2}\right)^{1/2}\\ &=-\left( \lim _{t \rightarrow 0^-}\frac{1-\cos t}{t^2}\right)^{1/2}\\ &=-\left(\frac{1}{2} \right)^{1/2}\\ &=-\frac{1}{\sqrt{2}}. \end{align*} Hence $ \lim _{t \rightarrow 0^+}\frac{\sqrt{1-\cos t}}{t}\neq \lim _{t \rightarrow 0^-}\frac{\sqrt{1-\cos t}}{t}$ and so the limit does not exist.
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