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Question Number 130837 by EDWIN88 last updated on 29/Jan/21

$$\underset{x\to 0}{\lim }\frac{{\sin }^{-1}\left(x\right)-{\tan }^{-1}\left(x\right)}{x^3}?$$

Answered by bemath last updated on 29/Jan/21

$$\left(1\right){\sin }^{-1}\left(\mathrm{x}\right)=\mathrm{x}+\frac{{\mathrm{x}}^3}{3!}+\frac{1^2.3^2{\mathrm{x}}^5}{5!}+...$$$$\left(2\right){\tan }^{-1}\left(\mathrm{x}\right)=\mathrm{x}-\frac{{\mathrm{x}}^3}{3}+\frac{{\mathrm{x}}^5}{5}-...$$$$\underset{x\to 0}{\lim }\frac{(\mathrm{x}+\frac{{\mathrm{x}}^3}{6})-(\mathrm{x}-\frac{{\mathrm{x}}^3}{3})}{{\mathrm{x}}^3}=\underset{x\to 0}{\lim }\frac{\frac{3}{6}{\mathrm{x}}^3}{{\mathrm{x}}^3}=\frac{1}{2}$$

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