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: 29


$$\lim_{x\to0}\frac{x+x\cos x}{\sin x\cos x}=2$$

Work Step by Step

$$A=\lim_{x\to0}\frac{x+x\cos x}{\sin x\cos x}$$ Multiply both numerator and denominator by $2$, we have $$A=\lim_{x\to0}\frac{2(x+x\cos x)}{2\sin x\cos x}$$ $$A=2\lim_{x\to0}\frac{x(1+\cos x)}{\sin2x}$$ $$A=2\lim_{x\to0}\frac{x}{\sin2x}\lim_{x\to0}(1+\cos x)$$ $$A=2\lim_{x\to0}\Big(\frac{\sin2x}{x}\Big)^{-1}(1+\cos0)$$ $$A=2\Big(\lim_{x\to0}\frac{\sin2x}{x}\Big)^{-1}(1+1)=2\Big(\lim_{x\to0}\frac{\sin2x}{x}\Big)^{-1}\times2$$ $$A=4\Big(\lim_{x\to0}\frac{\sin2x}{x}\Big)^{-1}=4X^{-1}$$ Considering $X$: $$X=\lim_{x\to0}\frac{\sin2x}{x}$$ Multiply both numerator and denominator by $2$: $$X=\lim_{x\to0}\frac{2\sin2x}{2x}=2\lim_{x\to0}\frac{\sin2x}{2x}$$ Apply Theorem 7 with $\theta=2x$: $$X=2\times1=2$$ Therefore, $$A=4X^{-1}=4\times(2)^{-1}=4\times\frac{1}{2}=2$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.