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Question Number 120823 by bramlexs22 last updated on 03/Nov/20

Answered by liberty last updated on 03/Nov/20

$$\rho =\underset{x\to 0}{\lim }\frac{\sqrt{\cos \mathrm{x}+\sqrt{2\sin {}^2\mathrm{x}+{\mathrm{x}}^3}}-1}{\mathrm{xsin}\mathrm{x}}$$$$\rho =\underset{x\to 0}{\lim }\frac{(\cos \mathrm{x}-1)+\sqrt{2\sin {}^2\mathrm{x}+{\mathrm{x}}^3}}{\mathrm{x}\sin \mathrm{x}}.\left(\frac{1}{2}\right)$$$$\rho =\frac{1}{2}[\underset{x\to 0}{\lim }\frac{-2\sin {}^2\left(\mathrm{x}/2\right)}{\mathrm{x}\sin \mathrm{x}}+\underset{x\to 0}{\lim }\sqrt{\frac{2\sin {}^2\mathrm{x}+{\mathrm{x}}^3}{{\mathrm{x}}^2\sin {}^2\mathrm{x}}}]$$$$\mathrm{the}\mathrm{second}\mathrm{limit}\underset{x\to 0}{\lim }\sqrt{\frac{2\sin {}^2\mathrm{x}+{\mathrm{x}}^3}{{\mathrm{x}}^2\sin {}^2\mathrm{x}}}=$$$$\underset{x\to 0}{\lim }\sqrt{\frac{{\mathrm{x}}^2(2+\mathrm{x})}{{\mathrm{x}}^2.{\mathrm{x}}^2}}=\underset{x\to 0}{\lim }\sqrt{\frac{2+\mathrm{x}}{{\mathrm{x}}^2}}=\mathrm{\infty }$$$$[\mathrm{as}\mathrm{x}\to 0\wedge \sin \mathrm{x}\approx \mathrm{x}]$$

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