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Question Number 143255 by Mathspace last updated on 12/Jun/21

$$find{lim}_{x\to 0}\frac{sin(1-cosx)+1-cos\left(sinx\right)}{x^2}$$

Answered by bramlexs22 last updated on 12/Jun/21

$$\underset{x\to 0}{\lim }\frac{\frac{\sin (1-\cos x)}{1-\cos x}.(1-\cos x)+1-\cos \left(\sin x\right)}{x^2}$$$$=\underset{x\to 0}{\lim }\frac{1-\cos x+1-\cos \left(\sin x\right)}{x^2}$$$$=\underset{x\to 0}{\lim }\frac{2\sin {}^2\left(\frac{x}{2}\right)+2\sin {}^2\left(\frac{\sin x}{2}\right)}{x^2}$$$$=\underset{x\to 0}{\lim }2{\left(\frac{\sin \frac{x}{2}}{x}\right)}^2+\underset{x\to 0}{\lim 2}{\left(\frac{\sin \left(\frac{\sin x}{2}\right)}{x}\right)}^2$$$$=2\times \frac{1}{4}+\underset{x\to 0}{\lim 2}{\left(\frac{\sin x}{2}\right)}^2$$$$=\frac{1}{2}$$

Answered by mathmax by abdo last updated on 12/Jun/21

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

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