\(f(x) = \lim\limits_{n \to \infty} \frac{x}{1 + (2\,sin\,x)^{2n}}\)
\(=\lim\limits_{n \to \infty} \frac{x}{1 + (4\,sin^2\,x)^n}\,...(1)\)
If \(4\,sin^2x = 1\)
\(\Rightarrow sin^2x = \frac{1}{4}\)
\(\Rightarrow sin\, x = \pm \frac{1}{2}\) \(= sin \left(\pm \frac{\pi}{6}\right)\)
\(\Rightarrow x = n\pi \pm \frac{\pi}{6}\)
If \(4\,sin^2x = 1\) then \(\lim\limits_{n \to \infty} \frac{x}{1 + (4\,sin^2\,x)^n}\)
\(=\lim\limits_{n \to \infty} \frac{x}{1 + 1^n}\)
\(=\lim\limits_{n \to \infty} \frac{x}{1 + 1}\) \((\because 1^n = 1)\)
\(=\frac{x}{x + 1} = \frac{x}{2}\)
If \(0 \leq 4\,sin^2x < 1\) then \(\lim\limits_{n \to \infty} (4\,sin^2x)^n = 0.\)
and \(f(x) = \lim\limits_{n \to \infty} \frac{x}{1 + (4\,sin^2\,x)^n}\)
\(=\frac{x}{1 + 0} = x\)
Hence, \(f(x) = \begin{cases} x; & \quad 0 \leq 4\,sin^2x < 1\\ \frac{x}{2}; & \quad 4\sin^2x = 1 \end{cases}\)
\(= \begin{cases} x; & \quad x \neq n\pi\, \pm \frac{\pi}{6},\,n \in I\\ \frac{x}{2}; & \quad x = n \pi \pm \frac{\pi}{6}, n \in I \end{cases}\)
\(\therefore\) f(x) is discontinuous at x = \(n\,\pi\,\pm \frac{\pi}{6},n\in I.\)
(∵ x ≠ \(\frac{x}{2}\) except x = 0)
