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Question Number 121813 by bemath last updated on 12/Nov/20

$$\underset{\theta \to 0}{\lim }\frac{\sin \theta -\theta \cos \theta }{\sin \theta -\theta }=?$$

Answered by bobhans last updated on 12/Nov/20

$$\underset{\theta \to 0}{\lim }\frac{\sin \theta -\theta \cos \theta }{\sin \theta -\theta }=$$$$\underset{\theta \to 0}{\lim }\frac{\theta -\frac{{\theta }^3}{6}-\theta (1-\frac{{\theta }^2}{2})}{\theta -\frac{{\theta }^3}{6}-\theta }=$$$$\underset{\theta \to 0}{\lim }\frac{\left(\frac{2{\theta }^3}{6}\right)}{-\left(\frac{{\theta }^3}{6}\right)}=-2.$$

Answered by liberty last updated on 12/Nov/20

$$\underset{\theta \to 0}{\lim }\frac{\cos \theta -(\cos \theta -\theta \sin \theta )}{\cos \theta -1}=$$$$\underset{\theta \to 0}{\lim }\frac{\theta \sin \theta }{-2\sin {}^2\left(\theta /2\right)}=\frac{1}{-2\left(\frac{1}{4}\right)}=-2.▴$$

Answered by Dwaipayan Shikari last updated on 12/Nov/20

$$\underset{x\to 0}{\lim }\frac{x-\frac{x^3}{3!}-x+\frac{x^3}{2!}}{x-\frac{x^3}{6}-x}=\underset{x\to 0}{\lim }\frac{\frac{x^3}{3}}{-\frac{x^3}{6}}=-2$$

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