Calculus (3rd Edition)

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

Chapter 2 - Limits - Chapter Review Exercises - Page 94: 42


The limit does not exist.

Work Step by Step

By substitution, we get the intermediate form $\frac{0}{0}$, so we have $$ \lim _{t \rightarrow 0} \frac{\sin^2 t}{t^2 } \frac{1}{t}=\lim _{t \rightarrow 0} \frac{\sin^2 t}{t^2 } \lim _{t \rightarrow 0} \frac{1}{t}\\ =(1)\frac{1}{0}. $$ We check the one-sided limits $$ \lim _{t \rightarrow 0^-} \frac{1}{t } =\frac{1}{0^-}=-\infty $$ $$ \lim _{t \rightarrow 0^+} \frac{1}{t } =\frac{1}{0^+}=\infty $$ so the limit does not exist.
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