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Question Number 100134 by bemath last updated on 25/Jun/20

$$\underset{x\to \frac{\pi }{2}}{\lim }\frac{4\sin \mathrm{x}-\sqrt{6\sqrt{\sin \mathrm{x}}+10}}{\frac{\pi }{2}-\mathrm{x}}?$$

Commented by bobhans last updated on 25/Jun/20

$$\mathrm{set}\frac{\pi }{2}-\mathrm{x}=\mathrm{t},\mathrm{x}=\frac{\pi }{2}-\mathrm{t}$$$$\underset{\mathrm{t}\to 0}{\lim }\frac{4\cos \mathrm{t}-\sqrt{6\sqrt{\cos \mathrm{t}}+10}}{\mathrm{t}}=$$$$\underset{\mathrm{t}\to 0}{\lim }\frac{16\cos {}^2\mathrm{t}-(6\sqrt{\cos \mathrm{t}}+10)}{8\mathrm{t}}=$$$$\underset{\mathrm{t}\to 0}{\lim }\frac{-16\sin 2\mathrm{t}-\frac{6(-\sin \mathrm{t})}{2\sqrt{\cos \mathrm{t}}}}{8}=0$$$$$$

Commented by Dwaipayan Shikari last updated on 25/Jun/20

$$\underset{x\to \frac{\pi }{2}}{\lim }\frac{-4\mathrm{cosx}-\frac{\frac{1}{2}.\frac{3}{\sqrt{\mathrm{sinx}}}\mathrm{cosx}}{\sqrt{6\sqrt{\mathrm{sinx}}+10}}}{-1}=0$$

Answered by mathmax by abdo last updated on 25/Jun/20

$$\mathrm{changement}\frac{\pi }{2}-\mathrm{x}=\mathrm{t}\mathrm{give}\mathrm{f}\left(\mathrm{x}\right)=\frac{4\mathrm{sinx}-\sqrt{6\sqrt{\mathrm{sinx}}+10}}{\frac{\pi }{2}-\mathrm{x}}$$$$=\frac{4\mathrm{cost}-\sqrt{6\sqrt{\mathrm{cost}}+10}}{\mathrm{t}}=\mathrm{g}\left(\mathrm{t}\right)\mathrm{we}\mathrm{have}\mathrm{cost}\sim 1-\frac{{\mathrm{t}}^2}{2}\Rightarrow \sqrt{\mathrm{cost}}\sim 1-\frac{{\mathrm{t}}^2}{4}\Rightarrow$$$$6\sqrt{\mathrm{cost}}\sim 6-\frac{3}{2}{\mathrm{t}}^2\Rightarrow \sqrt{6\sqrt{\mathrm{cost}}+10}\sim \sqrt{16-\frac{3}{2}{\mathrm{t}}^2}=4\sqrt{1-\frac{{3\mathrm{t}}^2}{32}}\sim 4(1-\frac{{3\mathrm{t}}^2}{64})\Rightarrow$$$$\mathrm{g}\left(\mathrm{t}\right)\sim \frac{4-{2\mathrm{t}}^2-4+\frac{{3\mathrm{t}}^2}{16}}{\mathrm{t}}=(-2+\frac{3}{16})\mathrm{t}\to 0\Rightarrow {\lim }_{\mathrm{x}\to \frac{\pi }{2}}\mathrm{f}\left(\mathrm{x}\right)=0$$

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