Question

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

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

Answer 1

Answer: (B)

$f(n)=\underset{x\to 0}{\lim }{e}^{\frac{1}{x}\left(\left(1+\sin \frac{x}{2}\right)\left(1+\sin \frac{x}{2^2}\right)\cdot \left(1+\sin \frac{x}{2^n}\right)-1\right)}$
$=\underset{x\to 0}{\lim }\frac{\left(1+\left(\sin \frac{x}{2}+\sin \frac{x}{2^2}+\dots +\sin \frac{x}{2^n}\right)+\left(\sin \frac{x}{2}\sin \frac{x}{2^2}+\dots \right)+\dots -1\right)}{x}$
$=\underset{x\to 0}{\lim }{e}^{\left(\frac{\sin \frac{x}{2}}{x}+\frac{\sin \left(\frac{x}{2^2}\right)}{x}+\dots +\sin \frac{\left(\frac{x}{2^n}\right)}{x}\right)}=e^{\left(\frac{1}{2}+\frac{1}{2^2}+\dots \frac{1}{2^n}\right)}$
$\therefore \underset{n\to \infty }{\lim }f(n)=e^{\frac{1/2}{1-\frac{1}{2}}}=e$