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Question Number 138996 by bramlexs22 last updated on 21/Apr/21

$$\underset{x\to \pi /2}{\lim }\frac{\sqrt{2}-\sqrt{1+\sin \mathrm{x}}}{\sqrt{2}\cos {}^2\mathrm{x}}=?$$$$$$

Answered by EDWIN88 last updated on 21/Apr/21

$$\underset{x\to \pi /2}{\lim }\frac{1-\sin \mathrm{x}}{\sqrt{2}(1-\sin \mathrm{x})}.\underset{x\to \pi /2}{\lim }\frac{1}{(1+\sin \mathrm{x})(\sqrt{2}+\sqrt{1+\sin \mathrm{x}})}$$$$=\frac{1}{\sqrt{2}}.\frac{1}{2\left(2\sqrt{2}\right)}=\frac{1}{8}$$

Answered by mathmax by abdo last updated on 22/Apr/21

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\frac{\sqrt{2}-\sqrt{1+\mathrm{sinx}}}{\sqrt{2}{\cos }^2\mathrm{x}}\Rightarrow \mathrm{f}\left(\mathrm{x}\right){=}_{\mathrm{x}=\frac{\pi }{2}-\mathrm{t}}\frac{\sqrt{2}-\sqrt{1+\mathrm{cost}}}{\sqrt{2}{\sin }^2\mathrm{t}}=\mathrm{g}\left(\mathrm{t}\right)$$$$(\mathrm{x}\to \frac{\pi }{2}\Rightarrow \mathrm{t}\to 0)\mathrm{cost}\sim 1-\frac{{\mathrm{t}}^2}{2}\Rightarrow 1+\mathrm{cost}\sim 2-\frac{{\mathrm{t}}^2}{2}\mathrm{and}\mathrm{sint}\sim \mathrm{t}$$$$\sqrt{1+\mathrm{cost}}\sim \sqrt{2-\frac{{\mathrm{t}}^2}{2}}=\sqrt{2}\sqrt{1-\frac{{\mathrm{t}}^2}{4}}\sim \sqrt{2}(1-\frac{{\mathrm{t}}^2}{8})\Rightarrow$$$$\mathrm{g}\left(\mathrm{t}\right)\sim \frac{\sqrt{2}-\sqrt{2}(1-\frac{{\mathrm{t}}^2}{8})}{\sqrt{2}{\mathrm{t}}^2}=\frac{1}{8}\Rightarrow {\lim }_{\mathrm{t}\to 0}\mathrm{g}\left(\mathrm{t}\right)=\frac{1}{8}={\lim }_{\mathrm{x}\to \frac{\pi }{2}}\mathrm{f}\left(\mathrm{x}\right)$$

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