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Question Number 132137 by benjo_mathlover last updated on 11/Feb/21

$$\underset{x\to 0}{\lim }{\left(\frac{2}{{\mathrm{x}}^3}({\sin }^{-1}\mathrm{x}-{\tan }^{-1}\mathrm{x})\right)}^{\frac{2}{{\mathrm{x}}^2}}=?$$

Answered by EDWIN88 last updated on 11/Feb/21

$$\mathrm{L}=\underset{x\to 0}{\lim }{\left(\frac{2}{{\mathrm{x}}^3}[(\mathrm{x}+\frac{{\mathrm{x}}^3}{6}+\frac{{\mathrm{x}}^5}{40})-(\mathrm{x}-\frac{{\mathrm{x}}^3}{3}+\frac{{\mathrm{x}}^5}{5})]\right)}^{\frac{2}{{\mathrm{x}}^2}}$$$$\mathrm{L}=\underset{x\to 0}{\lim }{\left(\frac{2}{{\mathrm{x}}^3}(\frac{{\mathrm{x}}^3}{2}-\frac{{7\mathrm{x}}^5}{40})\right)}^{\frac{2}{{\mathrm{x}}^2}}$$$$\mathrm{L}=\underset{x\to 0}{\lim }{(1-\frac{{7\mathrm{x}}^2}{20})}^{\frac{2}{{\mathrm{x}}^2}}={\mathrm{e}}^{\underset{x\to 0}{\lim }(-\frac{{14\mathrm{x}}^2}{{20\mathrm{x}}^2})}={\mathrm{e}}^{-\frac{7}{10}}$$

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