Question

Given that $\prod _{n=1}^{n}cos\frac{x}{2^n}=\frac{\sin x}{2^n\sin \left(\frac{x}{2^n}\right)}$ and $f\left(x\right)=\{\begin{array}{cc}{lim}_{n\to \infty }{\sum }_{n=1}^n\frac{1}{2^n}\tan \left(\frac{x}{2^n}\right),& x\in (0,\pi )-\left\{\frac{\pi }{2}\right\}\\ \frac{2}{\pi }& x=\frac{\pi }{2}\end{array}$
Then which one of the following is true?

• A. $f\left(x\right)$ has non-removable discontinuity of finite type at $x=\frac{\pi }{2}$.
• B. $f\left(x\right)$ has removable discontinuity at $x=\frac{\pi }{2}$
• C. $f\left(x\right)$ is continuous at $x=\frac{\pi }{2}$.
• D. $f\left(x\right)$ has non-removable discontinuity of infinite type at $x=\frac{\pi }{2}$

Answer 1

The correct option is C $f\left(x\right)$ is continuous at $x=\frac{\pi }{2}$.

We know that $\tan x=\frac{2\tan \left(\frac{x}{2}\right)}{1-{\tan }^2\left(\frac{x}{2}\right)}$

$\Rightarrow \cot x=\frac{1-{\tan }^2\left(\frac{x}{2}\right)}{2\tan \left(\frac{x}{2}\right)}=\frac{1}{2}\cot \left(\frac{x}{2}\right)-\frac{1}{2}\tan \left(\frac{x}{2}\right)$

$\Rightarrow \frac{1}{2}\tan \left(\frac{x}{2}\right)=\frac{1}{2}\cot \left(\frac{x}{2}\right)-\cot x\to 1$

Replacing x by $\frac{x}{2}$ and multiplying by $\frac{1}{2}$

$\frac{1}{4}\tan \left(\frac{x}{4}\right)=\frac{1}{4}\cot \left(\frac{x}{4}\right)-\frac{1}{2}\cot \left(\frac{x}{2}\right)$

Repeating the steps

$\frac{1}{2^n}\tan \left(\frac{x}{2^n}\right)=\frac{1}{2^n}\cot \left(\frac{x}{2^n}\right)-\frac{1}{2^{n-1}}\cot \left(\frac{x}{2^{n-1}}\right)$

$\therefore {\sum }_{n=1}^n\left[\frac{1}{2^n}\tan \left(\frac{x}{2^n}\right)\right]=[\frac{1}{2}\cot \left(\frac{x}{2}\right)-\cot x]+[\frac{1}{4}\cot \left(\frac{x}{4}\right)-\frac{1}{2}\cot \left(\frac{x}{2}\right)]+....+[\frac{1}{2^n}\cot \left(\frac{x}{2^n}\right)-\frac{1}{2^{n-1}}\cot \left(\frac{x}{2^{n-1}}\right)]$

$=\frac{1}{2^n}\cot \left(\frac{x}{2^n}\right)-\cot x=\frac{\left(\frac{1}{2^n}\right)}{\tan \left(\frac{x}{2^n}\right)}-\cot x$

$f\left(x\right)={lim}_{n\to \infty }{\sum }_{n=1}^n\left[\frac{1}{2^n}\tan \left(\frac{x}{2^n}\right)\right]={lim}_{n\to \infty }(\frac{\left(\frac{x}{2^n}\right)\times \frac{1}{x}}{\tan \left(\frac{x}{2^n}\right)}-\cot x)$

as $n\to \infty ,\frac{1}{2^n}\to 0$

$\frac{x}{2^n}\to 0$

$f\left(x\right)=\frac{1}{x}\times 1-\cot x$

$x=\frac{\pi }{2}$

$f\left(\frac{\pi }{2}\right)=\frac{2}{\pi }$ (given )${lim}_{x\to \frac{\pi }{2}}f\left(x\right)={lim}_{x\to \frac{\pi }{2}}(\frac{1}{x}-\cot x)$

$=\frac{1}{\frac{\pi }{2}}-0=\frac{2}{\pi }$

$\therefore f\left(\frac{\pi }{2}\right)={lim}_{x\to \frac{\pi }{2}}f\left(x\right)$

$\Rightarrow f\left(x\right)$ is continuous at $x=\frac{\pi }{2}$