Question

Given that $\prod _{n=1}^n\cos \frac{x}{2^n}=\frac{\sin x}{2^n\sin \left(\frac{x}{2^n}\right)}$.
Let $f(x)=\{\begin{array}{l}\underset{n\to \infty }{\mathrm{Lim}}\sum _{n=1}^n\frac{1}{2^n}\tan \left(\frac{x}{2^n}\right),x\in (0,\pi )-\left\{\frac{\pi }{2}\right\}\\ \frac{2}{\pi },x=\frac{\pi }{2}\end{array}$ Then which one of the following altemative is True?

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

Answer 1

Answer: (C)

Given that $\cos \frac{x}{2}\cos \frac{x}{2^2}\cos \frac{x}{2^3}\dots \dots \cos \frac{x}{2^n}=\frac{\sin x}{2^n\sin \left(\frac{x}{2^n}\right)}$ ....(1)
Taking logarithm to the base 'e' on both sides of equation (1) and then differentiating w.r.t. x, we get
$\sum _{n=1}^n\frac{1}{2^n}\tan \frac{x}{2^n}=\left(\frac{1}{2^n}\cot \frac{x}{2^n}-\cot x\right)$
$\therefore \underset{n\to \infty }{\mathrm{Lim}}\sum _{n=1}^n\frac{1}{2^n}\tan \frac{x}{2^n}=\underset{n\to \infty }{\mathrm{Lim}}\left(\frac{1}{x}\times \frac{\frac{x}{2^n}}{\tan \frac{x}{2^n}}-\cot x\right)=\left(\frac{1}{x}-\cot x\right)$
$\therefore$ we have $f(x)=\{\begin{array}{ll}\frac{1}{x}-\cot x,& x\in (0,\pi )-\left\{\frac{\pi }{2}\right\}\\ \frac{2}{\pi },& x=\frac{\pi }{2}\end{array}$
Clearly $\underset{x\to \frac{\pi }{2}}{\mathrm{Lim}}f(x)=\underset{x\to \frac{\pi }{2}}{\mathrm{Lim}}\left(\frac{1}{x}-\cot x\right)=\frac{2}{\pi }=f\left(\frac{\pi }{2}\right)$
Hence $f(x)$ is continuous at $x=\frac{\pi }{2}$