Question

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

Answer 1

$L=\underset{n\to \infty }{lim}f\left(n\right)$

$f\left(n\right)=\underset{x\to 0}{lim}\left\{\right(1+\sin \left(x/2\right)\left)\right(1+\sin \left(x/2^2\right))....(1+\sin \left(x/2^n\right)){\}}^{1/x}$

$\log f\left(n\right)=\underset{x\to 0}{lim}\left\{\frac{\log \left\{\right(1+\sin \left(x/2\right)\left)\right(1+\sin \left(x/2^2\right))....(1+\sin \left(x/2^n\right)\left)\right\}}{x}\right\}$

$=\underset{x\to 0}{lim}\{\frac{\log (1+\sin (x/2\left)\right)}{x}+\frac{\log (1+\sin (x/2^2\left)\right)}{x}+....+\frac{\log (1+\sin (x/2^n\left)\right)}{x}\}$

By L-Hospital's rule:

$=\underset{x\to 0}{lim}\{\frac{\cos \left(x/2\right)}{2(1+\sin (x/2\left)\right)}+\frac{\cos \left(x/2^2\right))}{2^2(1+sin(x/2^2\left)\right)}+....+\frac{\cos \left(x/2^n\right)}{2^n(1+\sin (x/2^n\left)\right)}\}$

$=\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^n}$

$⟹f\left(n\right)=e^{\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^n}}$

$⟹L=\underset{n\to \infty }{lim}{e}^{\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^n}}$

$=e^{\frac{1}{1-\frac{1}{2}}}$ [Infinite GP with ratio $\frac{1}{2}$]

$=e^2$