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Question Number 143677 by bobhans last updated on 17/Jun/21

Answered by bemath last updated on 17/Jun/21

Answered by mathmax by abdo last updated on 17/Jun/21

$$1)\mathrm{f}\left(\mathrm{x}\right)=\frac{1-\cos (1-\mathrm{cosx})}{{\mathrm{x}}^4}\mathrm{we}\mathrm{have}1-\mathrm{cosx}\sim \frac{{\mathrm{x}}^2}{2}\Rightarrow$$$$\cos (1-\mathrm{cosx})\sim \cos \left(\frac{{\mathrm{x}}^2}{2}\right)\sim 1-\frac{{\mathrm{x}}^4}{8}\Rightarrow 1-\cos (1-\mathrm{cosx})\sim \frac{{\mathrm{x}}^4}{8}\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)\sim \frac{{\mathrm{x}}^4}{{8\mathrm{x}}^4}=\frac{1}{8}\Rightarrow {\lim }_{\mathrm{x}\to 0}\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{8}$$

Answered by mathmax by abdo last updated on 17/Jun/21

$$2)\mathrm{g}\left(\mathrm{x}\right)={\left(\frac{\mathrm{sinx}}{\mathrm{x}}\right)}^{\frac{\mathrm{sinx}}{\mathrm{x}-\mathrm{sinx}}}\Rightarrow \mathrm{g}\left(\mathrm{x}\right)={\mathrm{e}}^{\frac{\mathrm{sinx}}{\mathrm{x}-\mathrm{sinx}}\log \left(\frac{\mathrm{sinx}}{\mathrm{x}}\right)}$$$$\mathrm{we}\mathrm{have}\mathrm{sinx}\sim \mathrm{x}-\frac{{\mathrm{x}}^3}{6}\Rightarrow \frac{\mathrm{sinx}}{\mathrm{x}}\sim 1-\frac{{\mathrm{x}}^2}{6}\Rightarrow \log \left(\frac{\mathrm{sinx}}{\mathrm{x}}\right)\sim -\frac{{\mathrm{x}}^2}{6}$$$$\frac{\mathrm{sinx}}{\mathrm{x}-\mathrm{sinx}}\sim \frac{\mathrm{x}-\frac{{\mathrm{x}}^3}{6}}{\frac{{\mathrm{x}}^3}{6}}=\frac{6\mathrm{x}-{\mathrm{x}}^3}{{\mathrm{x}}^3}=\frac{6}{{\mathrm{x}}^2}-1\Rightarrow$$$$\frac{\mathrm{sinx}}{\mathrm{x}-\mathrm{sinx}}\log \left(\frac{\mathrm{sinx}}{\mathrm{x}}\right)\sim (\frac{6}{{\mathrm{x}}^2}-1)(-\frac{{\mathrm{x}}^2}{6})=-1+\frac{{\mathrm{x}}^2}{6}\Rightarrow$$$$\mathrm{g}\left(\mathrm{x}\right)\sim {\mathrm{e}}^{-1+\frac{{\mathrm{x}}^2}{6}}\Rightarrow {\lim }_{\mathrm{x}\to 0}\mathrm{g}\left(\mathrm{x}\right)={\mathrm{e}}^{-1}=\frac{1}{\mathrm{e}}$$

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