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Question Number 133758 by EDWIN88 last updated on 24/Feb/21

$$\underset{x\to \mathrm{\infty }}{\lim }{x}^2\ln (1+\frac{1}{x^{2/3}})({\tan }^{-1}\sqrt[3]{x+1}-{\tan }^{-1}\sqrt[3]{x-1})=?$$

Answered by liberty last updated on 24/Feb/21

$$\left(\mathrm{i}\right)\ln (1+\frac{1}{{\mathrm{x}}^{2/3}})\approx \frac{1}{{\mathrm{x}}^{2/3}}$$$$\left(\mathrm{ii}\right){\tan }^{-1}(\sqrt[3]{\mathrm{x}+1}-{\tan }^{-1}\left(\sqrt[3]{\mathrm{x}-1}\right))={\tan }^{-1}\left(\frac{\sqrt[3]{\mathrm{x}+1}-\sqrt[3]{\mathrm{x}-1}}{1+\sqrt[3]{{\mathrm{x}}^2-1}}\right)$$$${\tan }^{-1}\left(\mathrm{x}\right)\approx \mathrm{x}$$$${\tan }^{-1}\left(\frac{\sqrt[3]{\mathrm{x}+1}-\sqrt[3]{\mathrm{x}-1}}{1+\sqrt[3]{{\mathrm{x}}^2-1}}\right)\approx \frac{2}{{3\mathrm{x}}^{2/3}(1+{\mathrm{x}}^{1/3})}=\frac{2}{{3\mathrm{x}}^{4/3}}$$$$(\bullet ){\underset{x\to \mathrm{\infty }}{\lim x}}^2\left(\frac{1}{{\mathrm{x}}^{2/3}}\right)\left(\frac{2}{{3\mathrm{x}}^{4/3}}\right)=\frac{2}{3}$$

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