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: 26


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

Work Step by Step

$$A=\lim_{t\to0}\frac{2t}{\tan t}=\lim_{t\to0}\Big(\frac{\tan t}{2t}\Big)^{-1}=\Big(\lim_{t\to0}\frac{\tan t}{2t}\Big)^{-1}=X^{-1}$$ Considering $X$: $$X=\lim_{t\to0}\frac{\tan t}{2t}$$ $$X=\lim_{t\to0}\frac{\frac{\sin t}{\cos t}}{2t}=\lim_{t\to0}\frac{\sin t}{2t\cos t}$$ $$X=\lim_{t\to0}\frac{\sin t}{2t}\lim_{t\to0}\frac{1}{\cos t}$$ $$X=\lim_{t\to0}\frac{\sin t}{2t}\times\Big(\frac{1}{\cos0}\Big)$$ $$X=\lim_{t\to0}\frac{\sin t}{2t}\times\Big(\frac{1}{1}\Big)$$ $$X=\lim_{t\to0}\frac{\sin t}{2t}$$ To be able to apply Theorem 7, we need $t$ in the denominator, instead of $2t$. So we would multiply both numerator and denominator by $1/2$: $$X=\lim_{t\to0}\frac{1/2\sin t}{1/2\times2t}=\frac{1}{2}\lim_{t\to0}\frac{\sin t}{t}$$ Apply Theorem 7: $$X=\frac{1}{2}\times1=\frac{1}{2}$$ Therefore, $$A=X^{-1}=\Big(\frac{1}{2}\Big)^{-1}=2$$
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