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Question Number 120427 by bramlexs22 last updated on 31/Oct/20

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

Answered by john santu last updated on 31/Oct/20

$$\underset{x\to 0}{\lim }\frac{\sqrt{\cos x+\sqrt{2\sin x+x}}-1}{\sin x}=$$$$\underset{x\to 0}{\lim }\frac{1}{\sqrt{\cos x+\sqrt{2\sin x+x}}+1}.\underset{x\to 0}{\lim }\frac{\cos x-1+\sqrt{2\sin x+x}}{\sin x}$$$$\frac{1}{2}.\underset{x\to 0}{\lim }\frac{-2\sin {}^2\left(x/2\right)+\sqrt{2\sin x+x}}{2\sin \left(x/2\right)\cos \left(x/2\right)}$$$$\frac{1}{2}.[\underset{x\to 0}{\lim }-\tan \left(\frac{x}{2}\right)+\underset{x\to 0}{\lim }\sqrt{\frac{2\sin x+x}{\sin {}^{2}x}}]$$$$=\frac{1}{2}.[0+\underset{x\to 0}{\lim }\sqrt{\frac{3}{x}}]=\mathrm{\infty }$$

Answered by mathmax by abdo last updated on 31/Oct/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\frac{\sqrt{\mathrm{cosx}+\sqrt{2\mathrm{sinx}+\mathrm{x}}}-1}{\mathrm{sinx}}\mathrm{we}\mathrm{have}$$$$\mathrm{cosx}\sim 1-\frac{{\mathrm{x}}^2}{2}\mathrm{and}2\sin \left(\mathrm{x}\right)+\mathrm{x}\sim 2(\mathrm{x}-\frac{{\mathrm{x}}^3}{6})+\mathrm{x}=3\mathrm{x}-\frac{{\mathrm{x}}^3}{3}\Rightarrow$$$$\sqrt{2\mathrm{sinx}+\mathrm{x}}\sim \sqrt{3\mathrm{x}-\frac{{\mathrm{x}}^3}{3}}=\sqrt{3\mathrm{x}}\sqrt{1-\frac{{\mathrm{x}}^2}{9}}\sim \sqrt{3\mathrm{x}}(1-\frac{{\mathrm{x}}^2}{18})\Rightarrow$$$$\mathrm{cosx}+\sqrt{2\mathrm{sinx}+\mathrm{x}}-1\sim 1-\frac{{\mathrm{x}}^2}{2}+\sqrt{3\mathrm{x}}(1-\frac{{\mathrm{x}}^2}{18})-1$$$$\sim -\frac{{\mathrm{x}}^2}{2}+\sqrt{3\mathrm{x}}-\frac{{\mathrm{x}}^2\sqrt{3\mathrm{x}}}{18}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)\sim -\frac{\mathrm{x}}{2}+\frac{\sqrt{3}}{\sqrt{\mathrm{x}}}-\mathrm{x}\sqrt{\mathrm{x}}\sqrt{3}\Rightarrow$$$${\lim }_{\mathrm{x}\to 0^{+}}\mathrm{f}\left(\mathrm{x}\right)=+\mathrm{\infty }$$

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