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Question Number 126685 by bramlexs22 last updated on 23/Dec/20

$$\underset{x\to 0}{\lim }\frac{1-\cos x\sqrt{\cos 2x}}{x\sin x}=?$$

Answered by liberty last updated on 23/Dec/20

$$(\ast )\underset{x\to 0}{\lim }\frac{1-\cos x\sqrt{\cos 2x}}{x\sin x}=$$$$\underset{x\to 0}{\lim }\frac{1-(1-\frac{1}{2}{x}^2)\sqrt{1-\frac{1}{2}\left(4x^2\right)}}{x(x-\frac{x^3}{6})}=$$$$\underset{x\to 0}{\lim }\frac{1-(1-\frac{1}{2}{x}^2)(1-x^2)}{x^2(1-\frac{x^2}{6})}=$$$$\underset{x\to 0}{\lim }\frac{1-(1-\frac{3}{2}{x}^2+\frac{1}{2}{x}^4)}{x^2(1-\frac{x^2}{6})}=$$$$\underset{x\to 0}{\lim }\frac{x^2(\frac{3}{2}-\frac{1}{2}{x}^2)}{x^2(1-\frac{x^2}{6})}=\underset{x\to 0}{\lim }\frac{\frac{3}{2}-\frac{x^2}{2}}{1-\frac{x^2}{6}}=\frac{3}{2}$$

Answered by Dwaipayan Shikari last updated on 23/Dec/20

$$\underset{x\to 0}{\lim }\frac{1-{cos}^{2}xcos2x}{xsinx}.\frac{1}{1+cosx\sqrt{cos2x}}sinx\to x$$$$\underset{x\to 0}{\lim }\left(\frac{1-{cos}^{2}x+2{cosxsin}^{2}x)}{xsinx}\right).\frac{1}{2}=\left(\frac{x^2+2x^2}{x^2}\right).\frac{1}{2}=\frac{3}{2}$$

Answered by bramlexs22 last updated on 23/Dec/20

$$\underset{x\to 0}{\lim }\frac{-(-\sin x\sqrt{\cos 2x}+\cos x\left(\frac{-\sin 2x}{\sqrt{\cos 2x}}\right))}{\sin x+x\cos x}=$$$$\underset{x\to 0}{\lim }\frac{\sin x\cos 2x+\cos x\sin 2x}{\sqrt{\cos 2x}(\sin x+x\cos x)}=$$$$\underset{x\to 0}{\lim }\frac{\frac{\sin x}{x}.\cos 2x+\cos x.\frac{\sin 2x}{x}}{1(\frac{\sin x}{x}+\cos x)}=$$$$\underset{x\to 0}{\lim }\frac{\cos 2x+2\cos x}{1+\cos x}=\frac{3}{2}$$

Answered by mathmax by abdo last updated on 24/Dec/20

$$\mathrm{let}\mathrm{L}\left(\mathrm{x}\right)=\frac{1-\mathrm{cosx}\sqrt{\cos \left(2\mathrm{x}\right)}}{\mathrm{xsinx}}\mathrm{we}\mathrm{have}\mathrm{for}\mathrm{x}\sim 0$$$$\mathrm{sinx}\sim \mathrm{x},\mathrm{cosx}\sim 1-\frac{{\mathrm{x}}^2}{2},\cos \left(2\mathrm{x}\right)\sim 1-{2\mathrm{x}}^2\mathrm{and}\sqrt{\cos \left(2\mathrm{x}\right)}\sim \sqrt{1-{2\mathrm{x}}^2}$$$$\sim 1-{\mathrm{x}}^2\Rightarrow \mathrm{L}\left(\mathrm{x}\right)\sim \frac{1-(1-\frac{{\mathrm{x}}^2}{2})(1-{\mathrm{x}}^2)}{{\mathrm{x}}^2}=\frac{1-(1-{\mathrm{x}}^2-\frac{{\mathrm{x}}^2}{2}+\frac{{\mathrm{x}}^4}{2})}{{\mathrm{x}}^2}$$$$=\frac{\frac{3}{2}{\mathrm{x}}^2-\frac{{\mathrm{x}}^4}{2}}{{\mathrm{x}}^2}=\frac{3}{2}-\frac{{\mathrm{x}}^2}{2}\Rightarrow \mathrm{L}\left(\mathrm{x}\right)\sim \frac{3}{2}-\frac{{\mathrm{x}}^2}{2}\Rightarrow {\lim }_{\mathrm{x}\to 0}\mathrm{L}\left(\mathrm{x}\right)=\frac{3}{2}$$

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