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Question Number 123874 by john_santu last updated on 29/Nov/20

$$\left(1\right)\underset{\theta \to 0}{\lim }\frac{\theta -\sin \theta \cos \theta }{\tan \theta -\theta }$$$$\left(2\right)\underset{x\to 0}{\lim }\frac{\sin 3x-3x+x^2}{\sin x\sin 2x}$$

Answered by liberty last updated on 29/Nov/20

$$solvethisquestion$$$$\left(1\right)\underset{\theta \to 0}{\lim }\frac{\theta -\sin \theta \cos \theta }{\tan \theta -\theta }$$$$\left(2\right)\underset{x\to 0}{\lim }\frac{\sin 3x-3x+x^2}{\sin x\sin 2x}$$$$Solution:$$$$\left(1\right)by{L}^{\prime }Hopitalrule$$$$\underset{\theta \to 0}{\lim }\frac{\theta -\frac{1}{2}\sin 2\theta }{\tan \theta -\theta }=\underset{\theta \to 0}{\lim }\frac{1-\cos 2\theta }{\sec {}^2\theta -1}$$$$\underset{\theta \to 0}{\lim }\frac{\cos {}^2\theta \left(2\sin {}^2\left(\frac{\theta }{2}\right)\right)}{\sin {}^2\theta }=\underset{\theta \to 0}{\lim }2{\left(\frac{\sin \frac{\theta }{2}}{\tan \theta }\right)}^2=\frac{1}{2}$$$$\left(2\right)\underset{x\to 0}{\lim }\frac{\sin 3x-3x+x^2}{-\frac{1}{2}(\cos 3x-\cos x)}=$$$$-2\times \underset{x\to 0}{\lim }\frac{3\cos 3x-3+2x}{-3\sin 3x+\sin x}=-2\times \underset{x\to 0}{\lim }\frac{-9\sin 3x+2}{-9\cos 3x+\cos x}$$$$=-2\times \left[\frac{2}{-8}\right]=\frac{1}{2}.$$

Answered by Dwaipayan Shikari last updated on 29/Nov/20

$$1)\underset{x\to 0}{\lim }\frac{x-\frac{1}{2}sin2x}{x+\frac{x^3}{3}-x}=\frac{x-x+\frac{8}{12}{x}^3}{\frac{x^3}{3}}=2(sinx=x-\frac{x^3}{6})$$$$2)\underset{x\to 0}{\lim }\frac{3sinx-4{sin}^{3}x-3x+x^2}{x.2x}=\frac{3x-4x^3-3x+x^2}{2x^2}=\frac{1}{2}-2x$$$$=\frac{1}{2}(sinx\to x)$$

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