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: 25


$$\lim_{x\to0}\frac{\tan 2x}{x}=2$$

Work Step by Step

$$A=\lim_{x\to0}\frac{\tan 2x}{x}$$ $$A=\lim_{x\to0}\frac{\frac{\sin 2x}{\cos2x}}{x}=\lim_{x\to0}\frac{\sin 2x}{x\cos2x}$$ $$A=\lim_{x\to0}\frac{\sin2x}{x}\lim_{x\to0}\frac{1}{\cos2x}$$ $$A=\lim_{x\to0}\frac{\sin2x}{x}\times\Big(\frac{1}{\cos0}\Big)$$ $$A=\lim_{x\to0}\frac{\sin2x}{x}\times\Big(\frac{1}{1}\Big)$$ $$A=\lim_{x\to0}\frac{\sin2x}{x}$$ To be able to apply Theorem 7, we need $2x$ in the denominator, instead of $x$. So we would multiply both numerator and denominator by $2$: $$A=\lim_{x\to0}\frac{2\sin2x}{2\times x}=2\lim_{x\to0}\frac{\sin 2x}{2x}$$ Take $\theta=2x$. Then as $x\to0$, $\theta\to2\times 0=0$ $$A=2\lim_{\theta\to0}\frac{\sin\theta}{\theta}$$ Apply Theorem 7: $$A=2\times1=2$$
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