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Question Number 147685 by mathmax by abdo last updated on 22/Jul/21

$$\mathrm{calculate}{\lim }_{\mathrm{x}\to 0}\frac{\sin \left(2\mathrm{sinx}\right)}{{\mathrm{x}}^2}$$

Answered by liberty last updated on 23/Jul/21

$$\underset{x\to 0}{\lim }\frac{\sin \left(2\sin x\right)}{x^2}=$$$$\underset{x\to 0}{\lim }\frac{2\sin x-\frac{8\sin {}^{3}x}{6}}{x^2}=$$$$\frac{1}{6}\underset{x\to 0}{\lim }\frac{12\sin x-8\sin {}^{3}x}{x^2}=$$$$\frac{1}{6}\underset{x\to 0}{\lim }\frac{\sin x(12-8\sin {}^{2}x)}{x^2}=$$$$\frac{1}{6}\underset{x\to 0}{\lim }\frac{12-8\sin {}^{2}x}{x}=\mathrm{\infty }$$

Answered by mathmax by abdo last updated on 23/Jul/21

$$\mathrm{sinx}\sim \mathrm{x}-\frac{{\mathrm{x}}^3}{6}\Rightarrow 2\mathrm{sinx}\sim 2\mathrm{x}-\frac{{\mathrm{x}}^3}{3}\Rightarrow \sin \left(2\mathrm{sinx}\right)\sim \sin (2\mathrm{x}-\frac{{\mathrm{x}}^3}{3})$$$$\sim 2\mathrm{x}-\frac{{\mathrm{x}}^3}{3}-\frac{1}{6}{(2\mathrm{x}-\frac{{\mathrm{x}}^3}{3})}^3\Rightarrow \frac{\sin \left(2\mathrm{sinx}\right)}{{\mathrm{x}}^2}\sim \frac{2}{\mathrm{x}}-\frac{{\mathrm{x}}^2}{3}-\frac{\mathrm{x}}{6}{(2-\frac{{\mathrm{x}}^2}{3})}^3\Rightarrow$$$${\lim }_{\mathrm{x}\to 0}\frac{\sin \left(2\mathrm{sinx}\right)}{{\mathrm{x}}^2}=\stackrel{-}{+}\mathrm{\infty }$$

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