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Question Number 106810 by Study last updated on 07/Aug/20

$$li\underset{x\to 0}{m}\frac{tanx-sinx}{sinx(cos2x-cosx)}=???$$

Answered by bemath last updated on 07/Aug/20

$$@\mathrm{bemath}@$$$$\underset{x\to 0}{\lim }\frac{\tan \mathrm{x}-\sin \mathrm{x}}{\sin \mathrm{x}(\cos 2\mathrm{x}-\cos \mathrm{x})}?$$$$\underset{x\to 0}{\lim }\frac{\tan \mathrm{x}(1-\cos \mathrm{x})}{\sin \mathrm{x}(-2\sin \left(\frac{3\mathrm{x}}{2}\right)\sin \left(\frac{\mathrm{x}}{2}\right))}=$$$$\underset{x\to 0}{\lim }\frac{\tan \mathrm{x}(2\sin {}^2\left(\frac{\mathrm{x}}{2}\right)}{-2\sin \mathrm{x}\sin \left(\frac{3\mathrm{x}}{2}\right)\sin \left(\frac{\mathrm{x}}{2}\right)}=$$$$-1\times \underset{x\to 0}{\lim }\frac{\tan \mathrm{x}.\sin \left(\frac{\mathrm{x}}{2}\right)}{\sin \mathrm{x}\sin \left(\frac{3\mathrm{x}}{2}\right)}=-\frac{1}{3}$$

Answered by Dwaipayan Shikari last updated on 07/Aug/20

$$\underset{x\to 0}{\lim }\frac{sinx(\frac{1}{cosx}-1)}{sinx(-2sin3xsinx)}$$$$\underset{x\to 0}{\lim }\frac{\frac{2{sin}^2\frac{x}{2}}{cosx}}{-2sin\frac{3x}{2}sin\frac{x}{2}}=\underset{x\to 0}{\lim }\frac{\frac{2{\left(\frac{x}{2}\right)}^2}{cosx}}{-2.3x^2}.4=-\frac{1}{3}$$$$sinx\to x$$

Answered by 1549442205PVT last updated on 07/Aug/20

$$li\underset{x\to 0}{m}\frac{tanx-sinx}{sinx(cos2x-cosx)}=\underset{\mathrm{x}\to 0}{\lim }\frac{\mathrm{sinx}(\frac{1}{\mathrm{cosx}}-1)}{\mathrm{sinx}(\cos 2\mathrm{x}-\mathrm{cosx})}$$$$=li\underset{x\to 0}{m}\frac{\frac{1}{\mathrm{cosx}}-1}{cos2x-cosx)}=\underset{\mathrm{x}\to 0}{\lim }\frac{1-\mathrm{cosx}}{{2\cos }^2\mathrm{x}-\mathrm{cosx}-1}$$$$=\underset{\mathrm{x}\to 0}{\lim }\frac{1-\mathrm{cosx}}{(\mathrm{cosx}-1)(2\mathrm{cosx}+1)}=\underset{\mathrm{x}\to 0}{\lim }\frac{-1}{2\mathrm{cosx}+1}$$$$=-\frac{1}{3}$$

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