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Question Number 44795 by behi83417@gmail.com last updated on 04/Oct/18

Commented by maxmathsup by imad last updated on 05/Oct/18

$$letA\left(x\right)={\left(\frac{sinx}{x}\right)}^{\frac{1}{x^2}}\Rightarrow ln\left(A_{}\left(x\right)\right)=\frac{1}{x^2}ln\left(\frac{sinx}{x}\right)but$$$$sinx=x-\frac{x^3}{3!}+o\left(x^5\right)(x\to 0)\Rightarrow \frac{sinx}{x}=1-\frac{x^2}{6}+o\left(x^4\right)\Rightarrow$$$$ln\left(\frac{sinx}{x}\right)=ln(1-\frac{x^2}{6}+o\left(x^4\right))=-\frac{x^2}{6}+o\left(x^4\right)\Rightarrow \frac{1}{x^2}ln\left(\frac{sinx}{x}\right)=-\frac{1}{6}+o\left(x^2\right)\Rightarrow$$$${lim}_{x\to 0}A\left(x\right)=-\frac{1}{6}.$$

Commented by maxmathsup by imad last updated on 05/Oct/18

$${lim}_{x\to 0}ln\left(A\left(x\right)\right)=-\frac{1}{6}\Rightarrow {lim}_{x\to 0}A\left(x\right)=e^{-\frac{1}{6}}.$$

Commented by behi83417@gmail.com last updated on 05/Oct/18

$$thankyousomuchdearprop.abdo.$$

Commented by maxmathsup by imad last updated on 05/Oct/18

$$youarewelcomesirbehi.$$

Answered by ajfour last updated on 04/Oct/18

$$letthelimitbeL.$$$$\ln L=\underset{x\to 0}{\lim }\left\{\frac{1}{x^2}\ln (1-\frac{x^2}{6}+...)\right\}$$$$\Rightarrow L=e^{-1/6}.$$

Commented by behi83417@gmail.com last updated on 04/Oct/18

$$thankyouverymuchsirAjfour.$$

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