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: 27



Work Step by Step

$$A=\lim_{x\to0}\frac{x\csc2x}{\cos5x}$$ $$A=\lim_{x\to0}\frac{x\times\frac{1}{\sin2x}}{\cos5x}=\lim_{x\to0}\frac{x}{\sin2x\cos5x}$$ $$A=\lim_{x\to0}\frac{x}{\sin2x}\lim_{x\to0}\frac{1}{\cos5x}=\lim_{x\to0}\frac{x}{\sin2x}\times\frac{1}{\cos0}=\lim_{x\to0}\frac{x}{\sin2x}\times\frac{1}{1}$$ $$A=\lim_{x\to0}\frac{x}{\sin2x}$$ $$A=\lim_{x\to0}\Big(\frac{\sin2x}{x}\Big)^{-1}=\Big(\lim_{x\to0}\frac{\sin2x}{x}\Big)^{-1}=X^{-1}$$ Considering $X$: $$X=\lim_{x\to0}\frac{\sin2x}{x}$$ 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 $2$: $$X=\lim_{x\to0}\frac{2\sin 2x}{2\times x}=2\lim_{x\to0}\frac{\sin 2x}{2x}$$ Apply Theorem 7 with $\theta=2x$ here: $$X=2\times1=2$$ Therefore, $$A=X^{-1}=2^{-1}=\frac{1}{2}$$
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