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Question Number 148849 by EDWIN88 last updated on 31/Jul/21

$$\underset{x\to 0^{+}}{\lim }\frac{x^3-\cos \sqrt{x}\sin {}^{2}x}{x^2}=?$$

Answered by iloveisrael last updated on 01/Aug/21

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

Answered by mathmax by abdo last updated on 01/Aug/21

$${\lim }_{\mathrm{x}\to 0^{+}}\frac{{\mathrm{x}}^3-\cos \left(\sqrt{\mathrm{x}}\right){\sin }^2\mathrm{x}}{{\mathrm{x}}^2}={\lim }_{\mathrm{x}\to 0}\mathrm{x}-\cos \left(\sqrt{\mathrm{x}}\right){\left(\frac{\mathrm{sinx}}{\mathrm{x}}\right)}^2$$$$=0-\cos \left(0\right)\times {\left(1\right)}^2=-1$$

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