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Question Number 94730 by i jagooll last updated on 20/May/20

Answered by john santu last updated on 20/May/20

$$\underset{x\to 0^{+}}{\lim }{\mathrm{e}}^{\ln {\left(\mathrm{x}\right)}^{\sin \mathrm{x}}}=\underset{x\to 0}{\lim }{\mathrm{e}}^{\sin \mathrm{x}.\ln \left(\mathrm{x}\right)}$$$${\mathrm{e}}^{\underset{x\to 0^{+}}{\lim }\left(\sin \mathrm{x}\right).\ln \left(\mathrm{x}\right)}={\mathrm{e}}^{\underset{x\to 0^{+}}{\lim }\frac{\ln \left(\mathrm{x}\right)}{\csc \mathrm{x}}}$$$${\mathrm{e}}^{\underset{x\to 0^{+}}{\lim }\frac{\left(\frac{1}{\mathrm{x}}\right)}{\csc \left(\mathrm{x}\right).\cot \left(\mathrm{x}\right)}}={\mathrm{e}}^{\underset{x\to 0^{+}}{\lim }\frac{\left(\frac{\sin \mathrm{x}}{\mathrm{x}}\right)}{\cot \left(\mathrm{x}\right)}}$$$${\mathrm{e}}^{\underset{x\to 0^{+}}{\lim }\frac{\tan \left(\mathrm{x}\right).\sin \left(\mathrm{x}\right)}{\mathrm{x}}}={\mathrm{e}}^0=1$$

Commented by john santu last updated on 20/May/20

$$\mathrm{correct}?$$

Commented by i jagooll last updated on 21/May/20

$$\mathrm{yes}..\mathrm{correct}$$

Answered by mathmax by abdo last updated on 21/May/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^{\mathrm{sinx}}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)={\mathrm{e}}^{\mathrm{sinxln}\left(\mathrm{x}\right)}$$$$\mathrm{we}\mathrm{have}\mathrm{sinx}\mathrm{lnx}=\frac{\mathrm{sinx}}{\mathrm{x}}\times \left(\mathrm{xlnx}\right)\Rightarrow {\lim }_{\mathrm{x}\to 0^{+}}\mathrm{sinx}\mathrm{lnx}$$$$={\lim }_{\mathrm{x}\to 0^{+}}\frac{\mathrm{sinx}}{\mathrm{x}}\left(\mathrm{xlnx}\right)=1\times 0=0\Rightarrow {\lim }_{\mathrm{x}\to 0^{+}}\mathrm{f}\left(\mathrm{x}\right)=1$$

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