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Question Number 145491 by Mrsof last updated on 05/Jul/21

Commented by Mrsof last updated on 05/Jul/21

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Answered by puissant last updated on 05/Jul/21

$${\lim }_{\mathrm{x}\to 0}{\mathrm{e}}^{(\cos \left(\mathrm{x}\right)-1)\ln \left(\tan \left(\mathrm{x}\right)\right)}$$$$={\lim }_{\mathrm{x}\to 0}{\mathrm{e}}^{-\frac{{\mathrm{x}}^2}{2}\times \mathrm{x}}={\lim }_{\mathrm{x}\to 0}{\mathrm{e}}^{-\frac{{\mathrm{x}}^3}{2}}=1$$$$\mathrm{True}$$

Answered by mathmax by abdo last updated on 05/Jul/21

$$\mathrm{f}\left(\mathrm{x}\right)={\left(\mathrm{tanx}\right)}^{\mathrm{cosx}-1}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)={\mathrm{e}}^{(\mathrm{cosx}-1)\log \left(\mathrm{tanx}\right)}$$$$\mathrm{cosx}\sim 1-\frac{{\mathrm{x}}^2}{2}\Rightarrow \mathrm{cosx}-1\sim -\frac{{\mathrm{x}}^2}{2}\Rightarrow$$$$(\mathrm{cosx}-1)\log \left(\mathrm{tanx}\right)\sim -\frac{{\mathrm{x}}^2}{2}\log \left(\mathrm{tanx}\right)\sim -\frac{{\mathrm{x}}^2}{2}\mathrm{logx}\to 0\Rightarrow$$$${\lim }_{\mathrm{x}\to +\mathrm{\infty }}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{e}}^0=1$$

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