lim-x-0-sinx-x-tanx-x-

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

\lim_{x \to 0} \frac{\sin x - x}{\tan x - x} = ?

Answered by EDWIN88 last updated on 20/Mar/21

\lim_{x \to 0} \frac{\sin x - x}{\tan x - x} = \lim_{x \to 0} \frac{\cos x - 1}{\sec^{2} x - 1}

= \lim_{x \to 0} (\frac{\cos x - 1}{1 - \cos^{2} x}) . \lim_{x \to 0} \cos x

= - 1 . \lim_{x \to 0} \frac{\cos x - 1}{(\cos x - 1) (\cos x + 1)}

= - \frac{1}{2}

Answered by liberty last updated on 20/Mar/21

\lim_{x \to 0} \frac{\sin x - x}{\tan x - x}

= \lim_{x \to 0} \frac{x - \frac{x^{3}}{6} - x}{x + \frac{x^{3}}{3} - x} = - \frac{1}{2}

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