Question

If $\footnotesize\displaystyle f(n) = \lim _{x \to 0} \bigg \lbrace \!\!\!\bigg \lgroup1 + \sin \dfrac{x}{2} \bigg \rgroup \!\!\!\bigg \lgroup1 + \sin \dfrac{x}{ {2}^{2} } \bigg \rgroup \!\!\!\cdots\!\!\!\bigg \lgroup1 + \sin \dfrac{x}{ {2}^{n} } \bigg \rgroup\!\!\! \bigg \rbrace ^{ \frac{1}{x}}$ then find $\displaystyle \lim _{n \to \infty }f(n).$

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Answer 1

$\ \begin{aligned} \tt f(n) & \tt=\lim _{x \rightarrow 0} e^{\dfrac{1}{x}\left \lgroup\left(1+\sin \dfrac{x}{2}\right \rgroup\left(1+\sin \dfrac{x}{2^{2}}\right) \cdots\left(1+\sin \dfrac{x}{2^{n}}\right)-1\right)} \\ \\ & \tt=\lim _{x \rightarrow 0} e^{\dfrac{\left(1+\left(\sin \dfrac{x}{2}+\sin \dfrac{x}{2^{2}}+\ldots+\sin \dfrac{x}{2^{n}}\right)+\left(\sin \dfrac{x}{2} \sin \dfrac{x}{2^{2}}+\ldots\right)+\ldots-1\right)}{x}} \\ \\ & \tt=\lim _{x \rightarrow 0} e^{\left \lgroup\dfrac{\sin \dfrac{x}{2}}{x}+\dfrac{\sin \left(\dfrac{x}{2^{2}}\right)}{x}+\ldots+\sin \dfrac{\left(\dfrac{x}{2^{n}}\right)}{x}\right \rgroup}=e^{\left(\dfrac{1}{2}+\dfrac{1}{2^{2}}+\ldots \dfrac{1}{2^{n}}\right)} \\ \therefore \quad & \tt \lim _{n \rightarrow \infty} f(n)=e^{\dfrac{1 / 2}{1-\dfrac{1}{2}}}=e \end{aligned}$