## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 4 - Section 4.4 - Indeterminate Forms and l''Hospital''s Rule - 4.4 Exercises: 52

#### Answer

$$\lim_{x\to0}(\csc x-\cot x)=0$$

#### Work Step by Step

$$A=\lim_{x\to0}(\csc x-\cot x)$$ $$A=\lim_{x\to0}(\frac{1}{\sin x}-\frac{\cos x}{\sin x})$$ ($\csc x=\frac{1}{\sin x}$ and $\cot x=\frac{\cos x}{\sin x}$) $$A=\lim_{x\to0}\frac{1-\cos x}{\sin x}$$ Here there is no need for L'Hospital's Rule. Multiply both numerator and denominator by $(1+\cos x)$. We will try to use the identity: $1-\cos^2x=\sin^2 x.$ $$A=\lim_{x\to0}\frac{(1-\cos x)(1+\cos x)}{\sin x(1+\cos x)}$$ $$A=\lim_{x\to0}\frac{1-\cos^2 x}{\sin x(1+\cos x)}$$ $$A=\lim_{x\to0}\frac{\sin^2 x}{\sin x(1+\cos x)}$$ $$A=\lim_{x\to0}\frac{\sin x}{1+\cos x}$$ $$A=\frac{\sin0}{1+\cos0}$$ $$A=\frac{0}{1+1}=0$$

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