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Question Number 141124 by iloveisrael last updated on 16/May/21

Answered by bobhans last updated on 16/May/21

Answered by EDWIN88 last updated on 16/May/21

$$\underset{x\to 0}{\lim }\frac{1}{\cos \left(\sin \mathrm{x}\right)+\sqrt{{\mathrm{x}}^4-{\mathrm{x}}^2+1}}.\underset{x\to 0}{\lim }\frac{\cos {}^2\left(\sin \mathrm{x}\right)-1+{\mathrm{x}}^2-{\mathrm{x}}^4}{{\mathrm{x}}^4}$$$$=\frac{1}{2}.\underset{x\to 0}{\lim }\frac{-\sin {}^2\left(\sin \mathrm{x}\right)-{\mathrm{x}}^4+{\mathrm{x}}^2}{{\mathrm{x}}^4}$$$$=-\frac{1}{2}.\underset{x\to 0}{\lim }\frac{{\mathrm{x}}^4+\sin {}^2\left(\sin \mathrm{x}\right)-{\mathrm{x}}^2}{{\mathrm{x}}^4}$$$$=-\frac{1}{2}.\underset{x\to 0}{\lim }\frac{{\mathrm{x}}^4+{(\sin \mathrm{x}-\frac{\sin {}^3\mathrm{x}}{6})}^2-{\mathrm{x}}^2}{{\mathrm{x}}^4}$$$$=-\frac{1}{2}.\underset{x\to 0}{\lim }\frac{{\mathrm{x}}^4+\sin {}^2\mathrm{x}{(1-\frac{\sin {}^2\mathrm{x}}{6})}^2-{\mathrm{x}}^2}{{\mathrm{x}}^4}$$$$=-\frac{1}{2}.\underset{x\to 0}{\lim }\frac{{\mathrm{x}}^4+{(\mathrm{x}-\frac{{\mathrm{x}}^3}{6})}^2(1-\frac{\sin {}^2\mathrm{x}}{3})-{\mathrm{x}}^2}{{\mathrm{x}}^4}$$$$=-\frac{1}{2}.\underset{x\to 0}{\lim }\frac{{\mathrm{x}}^4+{\mathrm{x}}^2(1-\frac{{\mathrm{x}}^2}{3})(1-\frac{{\mathrm{x}}^2}{3})-{\mathrm{x}}^2}{{\mathrm{x}}^4}$$$$=-\frac{1}{2}.\underset{x\to 0}{\lim }\frac{{\mathrm{x}}^4+{\mathrm{x}}^2{(1-\frac{{\mathrm{x}}^2}{3})}^2-{\mathrm{x}}^2}{{\mathrm{x}}^4}$$$$=-\frac{1}{2}.\underset{x\to 0}{\lim }\frac{{\mathrm{x}}^4+{\mathrm{x}}^2(1-\frac{{2\mathrm{x}}^2}{3})-{\mathrm{x}}^2}{{\mathrm{x}}^4}$$$$=-\frac{1}{2}.\underset{x\to 0}{\lim }\frac{{\mathrm{x}}^4+\cancel{{\mathrm{x}}^2}-\frac{{2\mathrm{x}}^4}{3}-\cancel{{\mathrm{x}}^2}}{{\mathrm{x}}^4}=-\frac{1}{2}.\underset{x\to 0}{\lim }\frac{\frac{1}{3}{\mathrm{x}}^4}{{\mathrm{x}}^4}$$$$=-\frac{1}{6}\divideontimes$$

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