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Question Number 72218 by aliesam last updated on 26/Oct/19

Commented by turbo msup by abdo last updated on 26/Oct/19

$$letA\left(x\right)={\left(\frac{sinx}{x}\right)}^{\frac{1}{x^2}}\Rightarrow$$$$A\left(x\right)=e^{\frac{1}{x^2}ln\left(\frac{sinx}{x}\right)}$$$$wehavesinx=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n{x}^{2n+1}}{(2n+1)!}$$$$=x-\frac{x^3}{6}+o\left(x^5\right)\Rightarrow \frac{sinx}{x}=1-\frac{x^2}{6}+o\left(x^4\right)$$$$\Rightarrow ln\left(\frac{sinx}{x}\right)=ln(1-\frac{x^2}{6}+o\left(x^4\right))$$$$\sim -\frac{x^2}{6}(x\to 0)\Rightarrow \frac{1}{x^2}ln\left(\frac{sinx}{x}\right)\sim -\frac{1}{6}$$$$\Rightarrow {lim}_{x\to 0}A\left(x\right)=e^{-\frac{1}{6}}=\frac{1}{\left({}^6\sqrt{e}\right)}$$

Commented by aliesam last updated on 26/Oct/19

$$godblessyou$$

Commented by mathmax by abdo last updated on 26/Oct/19

$$youarewelcome.$$

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