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Evaluate ("lim")(xto0)((sinx)/x)^(((sin...

Evaluate `("lim")_(xto0)((sinx)/x)^(((sinx)/(x-sinx)))`

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Since, `underset(xto0)lim(sinx)/(x)=1`
and `" "underset(xto0)lim(sinx)/(x-sinx)=underset(xto0)lim(1)/(((x)/(sinx)-1))=(1)/(1-1)=oo`
we have `underset(xto0)lim((sinx)/(x))^(((sinx)/(x-sinx)))=e^(underset(xto0)lim((sinx)/(x)-1)((sinx)/((x-sinx))`
`=e^(underset(xto0)lim-(sinx)/(x))=e^(-1)=(1)/(e)`
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