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Question Number 145205 by imjagoll last updated on 03/Jul/21

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

Answered by EDWIN88 last updated on 03/Jul/21

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

Answered by mathmax by abdo last updated on 03/Jul/21

$$\mathrm{f}\left(\mathrm{x}\right)=\frac{1-\cos (2\left(\sin \left(\mathrm{sinx}\right)\right)}{{\mathrm{x}}^2}\mathrm{we}\mathrm{have}\mathrm{sinx}\sim \mathrm{x}\Rightarrow \sin \left(\mathrm{sinx}\right)\sim \mathrm{x}$$$$\Rightarrow \cos \left(2\sin \left(\mathrm{sinx}\right)\right)\sim \cos \left(2\mathrm{x}\right)\sim 1-\frac{{\left(2\mathrm{x}\right)}^2}{2}=1-{2\mathrm{x}}^2\Rightarrow$$$$1-\cos \left(2\sin \left(\mathrm{sinx}\right)\right)\sim {2\mathrm{x}}^2\Rightarrow \mathrm{f}\left(\mathrm{x}\right)\sim \frac{{2\mathrm{x}}^2}{{\mathrm{x}}^2}=2\Rightarrow {\lim }_{\mathrm{x}\to 0}\mathrm{f}\left(\mathrm{x}\right)=2$$

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