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Question Number 107034 by mathdave last updated on 08/Aug/20

Commented by kaivan.ahmadi last updated on 08/Aug/20

$$=\frac{\sqrt{2}\mid sin6\pi \mid }{\sqrt{1-cos\pi }}=\frac{0}{\sqrt{1-(-1)}}=\frac{0}{\sqrt{2}}=0$$

Commented by kaivan.ahmadi last updated on 08/Aug/20

$$sin6\pi =0andcos\pi =-1$$$$$$

Commented by kaivan.ahmadi last updated on 08/Aug/20

$$doyoumean{lim}_{x\to \pi }\frac{\sqrt{2}\mid sin6x\mid }{\sqrt{1+cosx}}?$$

Commented by mathdave last updated on 08/Aug/20

$$yah$$

Commented by 1549442205PVT last updated on 08/Aug/20

$$\mathrm{If}\mathrm{the}\mathrm{question}\mathrm{was}\mathrm{corrected}\mathrm{such}\mathrm{as}$$$$\mathrm{then}\mathrm{we}\mathrm{can}\mathrm{solve}\mathrm{follows}\mathrm{as}:$$$$\mathrm{L}{im}_{x\to \pi }\frac{\sqrt{2}\mid sin6x\mid }{\sqrt{1+cosx}}=\underset{\mathrm{x}\to \pi }{\lim }\frac{\sqrt{2}\mid \sin 6\mathrm{x}\mid }{\sqrt{{2\cos }^2\frac{\mathrm{x}}{2}}}$$$$=\underset{\mathrm{x}\to \pi }{\lim }\frac{\sqrt{2}\mid \sin 6\mathrm{x}\mid }{\sqrt{2}\mid \cos \frac{\mathrm{x}}{2}\mid }=\underset{\mathrm{x}\to \pi }{\lim }\mid \frac{2\sin 3\mathrm{xcos}3\mathrm{x}}{\cos \frac{\mathrm{x}}{2}}\mid$$$$=\underset{\mathrm{x}\to \pi }{\lim }\mid \frac{2(3\mathrm{sinx}-{4\sin }^3\mathrm{x})\cos 3\mathrm{x}}{\cos \frac{\mathrm{x}}{2}}\mid$$$$=\mathrm{l}\underset{\mathrm{x}\to \pi }{\mathrm{im}}\mid \frac{\mathrm{sinx}}{\cos \frac{\mathrm{x}}{2}}\mid .\mid (6-{8\sin }^2\mathrm{x})\cos 3\mathrm{x}\mid$$$$=\underset{\mathrm{x}\to \pi }{\lim }\mid 2\sin \frac{\mathrm{x}}{2}\mid .\mid (6-{8\sin }^2\mathrm{x})\cos 3\mathrm{x}\mid$$$$=2.\mid (6-8).(-1)\mid =4$$

Commented by john santu last updated on 09/Aug/20

$$\mid (6-8\sin {}^2\pi )\cos 3\pi \mid =$$$$\mid (6-0).(-1)\mid =6?$$

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