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Question Number 147738 by liberty last updated on 23/Jul/21

$$\underset{x\to 0}{\lim }\frac{\arctan x-\arcsin x}{\sin {}^{3}x}=?$$

Answered by gsk2684 last updated on 23/Jul/21

$$\underset{x\to 0}{\lim }\frac{{\tan }^{-1}x-{\sin }^{-1}x}{\sin {}^{3}x}$$$$\approx \underset{x\to 0}{\lim }\frac{{\tan }^{-1}x-{\sin }^{-1}x}{x^3}\to \frac{0}{0},apply{L}^{\prime }Hopitalrule$$$$=\underset{x\to 0}{\lim }\frac{\frac{d}{dx}\{{\tan }^{-1}x-{\sin }^{-1}x\}}{\frac{d}{dx}{x}^3}$$$$=\underset{x\to 0}{\lim }\frac{\frac{1}{1+x^2}-\frac{1}{\sqrt{1-x^2}}}{3x^2}$$$$=\underset{x\to 0}{\lim }\frac{\sqrt{1-x^2}-(1+x^2)}{(1+x^2)\sqrt{1-x^2}3x^2}$$$$=\frac{1}{3(1+0^2)\sqrt{1-0^2}}\underset{x\to 0}{\lim }\frac{{\left(\sqrt{1-x^2}\right)}^2-{(1+x^2)}^2}{x^2\{\sqrt{1-x^2}+1+x^2\}}$$$$=\frac{1}{3}\frac{1}{\{\sqrt{1-0^2}+1+0^2\}}\underset{x\to 0}{\lim }\frac{(1-x^2)-(1+2x^2+x^4)}{x^2}$$$$=\frac{1}{6}\underset{x\to 0}{\lim }\frac{-3x^2-x^4}{x^2}$$$$=\frac{1}{6}\underset{x\to 0}{\lim }(-3-x^2)$$$$=\frac{1}{6}(-3)=-\frac{1}{2}$$$$OR$$$$\underset{x\to 0}{\lim }\frac{{\tan }^{-1}x-{\sin }^{-1}x}{\sin {}^{3}x}$$$$=\underset{x\to 0}{\lim }\frac{(x-\frac{x^3}{3}+\frac{x^5}{5}-...)-(x+\frac{x^3}{3!}+\frac{3^2{x}^5}{5!}+...)}{x^3}$$$$=\underset{x\to 0}{\lim }\frac{(-\frac{1}{3}-\frac{1}{6})x^3+...}{x^3}=-\frac{1}{2}$$

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

$${\arctan }^{{}^{\prime }}\mathrm{x}=\frac{1}{1+{\mathrm{x}}^2}\sim 1-{\mathrm{x}}^2\Rightarrow \mathrm{arctanx}\sim \mathrm{x}-\frac{{\mathrm{x}}^3}{3}$$$$({\arcsin }^{{}^{\prime }}\mathrm{x}=\frac{1}{\sqrt{1-{\mathrm{x}}^2}}={\left(\sqrt{1-{\mathrm{x}}^2}\right)}^{-\frac{1}{2}}\sim 1+\frac{{\mathrm{x}}^2}{2}\Rightarrow \mathrm{arcsinx}\sim \mathrm{x}+\frac{{\mathrm{x}}^3}{6}\mathrm{and}\mathrm{sinx}\sim \mathrm{x}\Rightarrow$$$$\frac{\mathrm{arctanx}-\mathrm{arcsinx}}{{\sin }^3\mathrm{x}}\sim \frac{\mathrm{x}-\frac{{\mathrm{x}}^3}{3}-\mathrm{x}-\frac{{\mathrm{x}}^3}{6}}{{\mathrm{x}}^3}\sim \frac{-{3\mathrm{x}}^3}{{6\mathrm{x}}^3}=-\frac{1}{2}\Rightarrow$$$${\lim }_{\mathrm{x}\to 0}\frac{\mathrm{arctanx}-\mathrm{arcsinx}}{{\sin }^3\mathrm{x}}=-\frac{1}{2}$$

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