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Question Number 109733 by bobhans last updated on 25/Aug/20

$$\underset{x\to 0}{\lim }\frac{x-\tan x}{\sin x-x}?$$

Answered by Dwaipayan Shikari last updated on 25/Aug/20

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

Answered by 1549442205PVT last updated on 25/Aug/20

$$\lim \frac{x-\tan x}{\sin x-x}\underset{{\mathrm{L}}^{\prime }\mathrm{Hopital}}{=}\underset{\mathrm{x}\to 0}{\lim }\frac{1-(1+{\tan }^2\mathrm{x})}{\mathrm{cosx}-1}$$$$=\underset{\mathrm{x}\to 0}{\lim }\frac{-{\tan }^2\mathrm{x}}{\mathrm{cosx}-1}\underset{{\mathrm{L}}^{\prime }\mathrm{Hopital}}{=}\underset{\mathrm{x}\to 0}{\lim }\frac{-2\mathrm{tanx}(1+{\tan }^2\mathrm{x})}{-\mathrm{sinx}}$$$$=\underset{\mathrm{x}\to 0}{\mathbf{lim}}\frac{2(1+{\mathbf{tan}}^2\mathbf{x})}{\mathbf{cosx}}=\frac{2(1+0)}{1}=2$$

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