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Question Number 206095 by RoseAli last updated on 06/Apr/24

Answered by Frix last updated on 07/Apr/24

$$\underset{x\to 0}{\lim }\frac{x-\sin x}{x-\tan x}\stackrel{\left[{\mathrm{l}}^{\prime }\mathrm{H}\widehat{\mathrm{o}pital}\right]}{=}\underset{x\to 0}{\lim }\frac{\frac{d}{dx}[x-\sin x]}{\frac{d}{dx}[x-\tan x]}=$$$$=\underset{x\to 0}{\lim }\frac{1-\cos x}{-{\tan }^{2}x}$$$$\mathrm{Let}t=\tan \frac{x}{2};x\to 0\Rightarrow t\to 0$$$$\cos x=\frac{1-t^2}{1+t^2};\tan x=\frac{2t}{1-t^2}$$$$\underset{x\to 0}{\lim }\frac{1-\cos x}{-{\tan }^{2}x}=\underset{t\to 0}{\lim }\frac{\frac{2t^2}{1+t^2}}{-\frac{4t^2}{{(1-t^2)}^2}}=$$$$=\underset{t\to 0}{\lim }-\frac{{(1-t^2)}^2}{2(t^2+1)}=-\frac{1}{2}$$

Answered by MM42 last updated on 07/Apr/24

$$x-sinx\sim \frac{1}{6}{x}^3\mathrm{\&}x-tanx\sim -\frac{1}{3}{x}^3$$$$\Rightarrow {lim}_{x\to 0}\frac{\frac{1}{6}{x}^3}{-\frac{1}{3}{x}^3}=-\frac{1}{2}✓$$

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