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Question Number 190708 by mnjuly1970 last updated on 09/Apr/23

Answered by cortano12 last updated on 10/Apr/23

$$\mathrm{L}={\mathrm{e}}^{\underset{x\to 0}{\lim }(1+\sin 2\mathrm{x}-{\tan }^{-1}\left(2\mathrm{x}\right)-1).\frac{1}{{\mathrm{x}}^3}}$$$$\mathrm{L}={\mathrm{e}}^{\underset{x\to 0}{\lim }8.\left(\frac{\sin 2\mathrm{x}-{\tan }^{-1}\left(2\mathrm{x}\right)}{{\left(2\mathrm{x}\right)}^3}\right)}$$$$\mathrm{let}{\tan }^{-1}\left(2\mathrm{x}\right)=\mathrm{u}\Rightarrow 2\mathrm{x}=\tan \mathrm{u}$$$$\mathrm{L}={\mathrm{e}}^{8.\underset{\mathrm{u}\to \mathrm{o}}{\lim }\left(\frac{\sin \left(\tan \mathrm{u}\right)-\mathrm{u}}{\tan {}^3\mathrm{u}}\right)}$$$$\mathrm{L}={\mathrm{e}}^{8.\underset{\mathrm{u}\to \mathrm{o}}{\lim }\left(\frac{\tan \mathrm{u}-\mathrm{u}}{{\mathrm{u}}^3}\right)}$$$$\mathrm{L}={\mathrm{e}}^{8/3}$$

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