Question

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