Question

If $f(n)=\displaystyle\lim _{x \rightarrow 0}\left(\left(1+\sin \frac{x}{2}\right)\left(1+\sin \frac{x}{2^{2}}\right) \cdots\left(1+\sin \frac{x}{2^{n}}\right)\right)^{\frac{1}{x}}$ then $\displaystyle\lim _{n \rightarrow \infty} f(n)=$

• A. $1$
• B. $e$
• C. $0$
• D. $\infty$

Answer 1

Answer: (B)

$f(n)=\displaystyle\lim _{x \rightarrow 0} e^{\frac{1}{x}\left(\left(1+\sin \frac{x}{2}\right)\left(1+\sin \frac{x}{2^{2}}\right) \cdot\left(1+\sin \frac{x}{2^{n}}\right)-1\right)}$
$=\displaystyle\lim _{x \rightarrow 0} \frac{\left(1+\left(\sin \frac{x}{2}+\sin \frac{x}{2^{2}}+\ldots+\sin \frac{x}{2^{n}}\right)+\left(\sin \frac{x}{2} \sin \frac{x}{2^{2}}+\ldots\right)+\ldots-1\right)}{x}$
$=\displaystyle\lim _{x \rightarrow 0} e^{\left(\frac{\sin \frac{x}{2}}{x}+\frac{\sin \left(\frac{x}{2^{2}}\right)}{x}+\ldots+\sin \frac{\left(\frac{x}{2^{n}}\right)}{x}\right)}=e^{\left(\frac{1}{2}+\frac{1}{2^{2}}+\ldots \frac{1}{2^{n}}\right)}$
$\therefore \displaystyle\lim _{n \rightarrow \infty} f(n)=e^{\frac{1 / 2}{1-\frac{1}{2}}}=e$