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Question Number 99761 by Lekhraj last updated on 23/Jun/20

Commented by Dwaipayan Shikari last updated on 23/Jun/20

$$Ans=1$$

Commented by Dwaipayan Shikari last updated on 23/Jun/20

$$=e^{li\underset{x\to 0^{+}}{m}(x-1)sinx}=1$$

Answered by mathmax by abdo last updated on 23/Jun/20

$$\mathrm{let}\mathrm{L}\left(\mathrm{x}\right)={\mathrm{x}}^{\mathrm{sinx}}\Rightarrow \mathrm{L}\left(\mathrm{x}\right)={\mathrm{e}}^{\mathrm{sinxln}\left(\mathrm{x}\right)}\mathrm{but}\mathrm{sinx}\ln \left(\mathrm{x}\right)\sim \mathrm{xln}\left(\mathrm{x}\right)\to 0(\mathrm{x}\to 0^{+})\Rightarrow$$$${\lim }_{\mathrm{x}\to 0^{+}}\mathrm{L}\left(\mathrm{x}\right)=1$$

Commented by Lekhraj last updated on 23/Jun/20

$$\mathrm{Thanks}.$$

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