Question

Let $P_n=\prod _{k=1}^{n}cos(x.2^{-k})$ and $g\left(x\right)=\underset{n\to \infty }{lim}{P}_n$, then $\underset{x\to 0}{lim}g\left(x\right)$ is

• A. $0$
• B. $1$
• C. $-1$
• D. does not exist

Answer 1

The correct option is B $1$

$P_n=\prod _{k=1}^n\cos (x.2^{-k})$

$=\cos \left(\frac{x}{2}\right)\cos \left(\frac{x}{4}\right)\dots \cos \left(\frac{x}{2^n}\right)$

$P_n=\prod _{k=1}^n\cos (x.2^{-k})=\frac{\sin \left(x\right)}{2\sin \left(\frac{x}{2}\right)}\frac{\sin \left(\frac{x}{2}\right)}{2\sin \left(\frac{x}{4}\right)}\frac{\sin \left(\frac{x}{4}\right)}{2\sin \left(\frac{x}{8}\right)}\dots \frac{\sin \left(\frac{x}{2^{n-1}}\right)}{2\sin \left(\frac{x}{2^n}\right)}$.....................$\because (\sin x=2\sin \frac{x}{2}\cos \frac{x}{2})$

$P_n=\prod _{k=1}^n\cos (x.2^{-k})=\frac{\sin \left(x\right)}{2^n\sin \left(\frac{x}{2^n}\right)}$

$g\left(x\right)={lim}_{n\to \infty }{P}_n={lim}_{n\to \infty }\frac{\sin \left(x\right)}{2^n\sin \left(\frac{x}{2^n}\right)}$

$g\left(x\right)={lim}_{n\to \infty }{P}_n={lim}_{n\to \infty }\frac{\sin \left(x\right)}{x\cdot \frac{\sin \left(\frac{x}{2^n}\right)}{\frac{x}{2^n}}}=\frac{\sin x}{x}$

${lim}_{x\to 0}g\left(x\right)={lim}_{x\to 0}\frac{\sin x}{x}=1$