University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Section 2.4 - One-Sided Limits - Exercises - Page 85: 24



Work Step by Step

$$A=\lim_{h\to0^-}\frac{h}{\sin3h}=\lim_{h\to0^-}\Big(\frac{\sin3h}{h}\Big)^{-1}=\Big(\lim_{h\to0^-}\frac{\sin3h}{h}\Big)^{-1}=X^{-1}$$ Consider $X$: $$X=\lim_{h\to0^-}\frac{\sin3h}{h}$$ To be able to apply Theorem 7, we need $3h$ in the denominator, instead of $h$. So we would multiply both numerator and denominator by $3$: $$X=\lim_{h\to0}\frac{3\sin3h}{3\times h}=3\lim_{h\to0}\frac{\sin 3h}{3h}$$ Take $\theta=3h$. Then as $h\to0$, $\theta\to3\times 0=0$ $$X=3\lim_{\theta\to0}\frac{\sin\theta}{\theta}$$ Apply Theorem 7: $$X=3\times1=3$$ Therefore, $$A=X^{-1}=3^{-1}=\frac{1}{3}$$
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