Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 4 - Applications of the Derivative - 4.7 L'Hopital's Rule - 4.7 Exercises: 47

Answer

The solution is $$\lim_{x\to0}\csc6x\sin7x=\frac{7}{6}.$$

Work Step by Step

To calculate this limit we will use L'Hopital's rule. "LR" will stand for "Apply L'Hopital's rule". $$\lim_{x\to0}\csc6x\sin7x=\lim_{x\to0}\frac{1}{\sin 6x}\cdot \sin7x=\lim_{x\to0}\frac{\sin7x}{\sin 6x}=\left[\frac{\sin (7\cdot0)}{\sin(6\cdot0)}\right]=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to0}\frac{(\sin7x)'}{(\sin6x)'}=\lim_{x\to0}\frac{\cos7x(7x)'}{\cos6x(6x)'}=\lim_{x\to0}\frac{7\cos7x}{6\cos6x}=\frac{7\cos(7\cdot0)}{6\cos(6\cdot0)}=\frac{7}{6}.$$
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