If `f(n)=underset(xto0)lim{(1+"sin"(x)/(2))(1+"sin"(x)/(2^(2)))...(1+"sin"(x)/(2^(n)))}^((1)/(x))` then find `underset(ntooo)limf(n).`

Explanation

`f(n)=underset(xto0)lime^((1)/(x){(1+"sin"(x)/(2))(1+"sin"(x)/(2^(2)))...(1+"sin"(x)/(2^(n)))-1})`
`=underset(xto0)lime^(({1+("sin"(x)/(2)+"sin"(x)/(2^(2))+...+"sin"(x)/(2^(n)))+("sin"(x)/(2)"sin"(x)/(2^(2))+...)+...-1})/(x))`
`=underset(xto0)lime^({("sin"(x)/(2))/(x)+(sin((x)/(2^(2))))/(x)+...+(sin((x)/(2^(n))))/(x)}`
`=e^(((1)/(2)+(1)/(2^(2))+...+(1)/(2^(n)))`
`:." "underset(ntooo)limf(n)=(1//2)/(e^(1-(1)/(2)))=e`