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Question Number 62539 by aliesam last updated on 22/Jun/19

$$\mathrm{li}\underset{\mathrm{x}\to \mathrm{\infty }}{\mathrm{m}}\frac{{\mathrm{x}}^3+\sin \left(2\mathrm{x}\right)-2\sin \left(\mathrm{x}\right)}{\arctan \left({\mathrm{x}}^3\right)-{\left(\arctan \left(\mathrm{x}\right)\right)}^3}$$

Answered by tanmay last updated on 22/Jun/19

$$\underset{x\to 0}{\lim }\frac{x^3+(2x-\frac{8x^3}{3!}+\frac{32x^5}{5!})-2(x-\frac{x^3}{3!}+\frac{x^5}{5!})}{(x^3-\frac{x^9}{3})-{(x-\frac{x^3}{3})}^3}$$$$\underset{x\to 0}{\lim }\frac{x^3-\frac{8x^3}{3!}+\frac{2x^3}{3!}+\frac{32x^5}{5!}-\frac{2x^5}{5!}}{x^3-\frac{x^9}{3}-x^3+3x^2\times \frac{x^3}{3}-3x\times \frac{x^6}{9}+\frac{x^9}{27}}$$$$\underset{x\to 0}{\lim }\frac{x^3(1+\frac{2}{3!}-\frac{8}{3!})+\frac{x^5}{5!}(32-2)}{\frac{-x^9}{3}+x^5-\frac{x^7}{3}+\frac{x^9}{27}}$$$$\underset{x\to 0}{\lim }\frac{\frac{30}{120}}{\frac{-x^4}{3}+1-\frac{x^2}{3}+\frac{x^4}{27}}=\frac{1}{4}\left[deviding{N}_{r}and{D}_{r}by{x}^5\right]$$$$$$$$ithink\underset{x\to 0}{\lim }not\underset{x\to \mathrm{\infty }}{\lim }$$$$$$$$$$$$$$$$\underset{x\to \mathrm{\infty }}{\mathrm{li}}$$$$$$

Commented by aliesam last updated on 22/Jun/19

$$\mathrm{yes}\mathrm{sir}\mathrm{its}\mathrm{typo}\mathrm{so}\mathrm{sorry}$$

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