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Question Number 136243 by Khalmohmmad last updated on 19/Mar/21

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

Answered by EDWIN88 last updated on 20/Mar/21

$$\underset{x\to 0}{\lim }\frac{\sin \mathrm{x}-\mathrm{x}}{\tan \mathrm{x}-\mathrm{x}}=\underset{x\to 0}{\lim }\frac{\cos \mathrm{x}-1}{\sec {}^2\mathrm{x}-1}$$$$=\underset{x\to 0}{\lim }\left(\frac{\cos \mathrm{x}-1}{1-\cos {}^2\mathrm{x}}\right).\underset{x\to 0}{\lim }\cos \mathrm{x}$$$$=-1.\underset{x\to 0}{\lim }\frac{\cos \mathrm{x}-1}{(\cos \mathrm{x}-1)(\cos \mathrm{x}+1)}$$$$=-\frac{1}{2}$$

Answered by liberty last updated on 20/Mar/21

$$\underset{x\to 0}{\lim }\frac{\sin x-x}{\tan x-x}$$$$=\underset{x\to 0}{\lim }\frac{x-\frac{x^3}{6}-x}{x+\frac{x^3}{3}-x}=-\frac{1}{2}$$

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