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Question Number 144645 by mathdanisur last updated on 27/Jun/21

$$\underset{x\to 0}{lim}{\left(\frac{1+tanx}{1+sinx}\right)}^{\frac{1}{\mathit{s}\mathit{i}\mathit{n}\mathit{x}}}=?$$

Answered by liberty last updated on 27/Jun/21

$$\underset{x\to 0}{\lim }{\left(\frac{1+\sin \mathrm{x}+\tan \mathrm{x}-\sin \mathrm{x}}{1+\sin \mathrm{x}}\right)}^{\frac{1}{\sin \mathrm{x}}}$$$$=\underset{x\to 0}{\lim }{(1+\frac{\tan \mathrm{x}-\sin \mathrm{x}}{1+\sin \mathrm{x}})}^{\frac{1}{\sin \mathrm{x}}}$$$$=\underset{x\to 0}{\lim }{\left[{(1+\frac{\tan \mathrm{x}-\sin \mathrm{x}}{1+\sin \mathrm{x}})}^{\frac{1+\sin \mathrm{x}}{\tan \mathrm{x}-\sin \mathrm{x}}}\right]}^{\frac{\tan \mathrm{x}-\sin \mathrm{x}}{\sin \mathrm{x}(1+\sin \mathrm{x})}}$$$$={\mathrm{e}}^{\underset{x\to 0}{\lim }\left(\frac{\tan \mathrm{x}-\sin \mathrm{x}}{\sin \mathrm{x}(1+\sin \mathrm{x})}\right)}$$$$={\mathrm{e}}^{\underset{x\to 0}{\lim }\left(\frac{\tan \mathrm{x}(1-\cos \mathrm{x})}{\sin \mathrm{x}(1+\sin \mathrm{x})}\right)}$$$$={\mathrm{e}}^{\frac{0}{1}}={\mathrm{e}}^0=1.$$

Commented by mathdanisur last updated on 27/Jun/21

$$alotcoolthankyouSir$$

Answered by mathmax by abdo last updated on 27/Jun/21

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

Commented by mathdanisur last updated on 27/Jun/21

$$ThanksSiralotcool$$

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