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Question Number 105643 by john santu last updated on 30/Jul/20

$$\underset{x\to 0}{\lim }\frac{\tan (x-\sin x)}{2-\sqrt{4+\sin {}^{3}2x}}$$

Answered by bramlex last updated on 30/Jul/20

$$\underset{x\to 0}{\lim }\frac{\tan (x-(x-\frac{1}{6}{x}^3))}{2-2\sqrt{1+\frac{\sin {}^{3}2x}{4}}}=$$$$\underset{x\to 0}{\lim }\frac{\tan \left(\frac{1}{6}{x}^3\right)}{2(1-(1+\frac{\sin {}^{3}2x}{8}))}=$$$$\underset{x\to 0}{\lim }\frac{\tan \left(\frac{1}{6}{x}^3\right)}{-\left(\frac{8}{4}{x}^3\right)}=-\frac{1}{6}\times \frac{1}{2}=-\frac{1}{12}▴$$

Answered by Dwaipayan Shikari last updated on 30/Jul/20

$$\underset{x\to 0}{\lim }\frac{tan(x-x+\frac{x^3}{6})}{4-4-{sin}^{3}2x}(2+\sqrt{4+{sin}^{3}2x})$$$$\underset{x\to 0}{\lim }\frac{tan\frac{x^3}{6}}{-{\left(2x\right)}^3}(2+\sqrt{4+8x^3})=\underset{x\to 0}{\lim }\frac{\frac{x^3}{6}}{-8x^3}(2+\sqrt{4})=-\frac{1}{12}\{{\left(sin2x\right)}^3\to 8x^3$$

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