91Ó°ÊÓ

First, we need to verify whether the given limit holds or not. We have: \(f(x) = \lim_{n \rightarrow \infty} (1 - \sin^2 x)^n = \lim_{n\rightarrow \infty}\cos^{2n}x\) Now, using the double angle formula: \(\cos 2x = 1 - 2\sin^2 x\). Rearrange it as \(\sin^2 x = \frac{1 - \cos 2x}{2}\). Substitute this into the expression, we get: \(f(x) = \lim_{n \rightarrow \infty} \left(1 - \frac{1 - \cos2x}{2}\right)^n\) So, it simplifies to: \(f(x) = \lim_{n \rightarrow \infty} \left(\frac{1 + \cos2x}{2}\right)^n\) Now, using the double angle formula for cosine: \(\cos^2 x = \frac{1+\cos 2x}{2}\). Replace \(\cos^2 x\) in the expression, and we get: \(f(x) = \lim_{n \rightarrow \infty} \cos^{2n}x\) Thus, the limit holds.

Use the expression obtained in step 1 and plug in \(x = \frac{\pi}{4}\): \(f\left(\frac{\pi}{4}\right) = \lim_{n \rightarrow \infty} \cos^{2n}\left(\frac{\pi}{4}\right)\) Since \(\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}\), we have: \(f\left(\frac{\pi}{4}\right) = \lim_{n \rightarrow \infty} \left(\frac{1}{\sqrt{2}}\right)^{2n} = \lim_{n \rightarrow \infty} \left(\frac{1}{2}\right)^n = 0\)

The function is given as discontinuous at \(x = n \pi\). Let's analyze the behavior of the function around these points: \(f(n\pi) = \lim_{n\rightarrow \infty}\cos^{2n}(2n\pi)\) Since \(\cos k\pi = (-1)^k\) for any integer k, we get: \(f(n\pi) = \lim_{n\rightarrow \infty} ((-1)^{2n})^{2n} = \lim_{n\rightarrow \infty} (1)^{2n} = 1\) It can be observed that as \(x\) approaches \(n\pi\), the function approaches its limit value of 1 but does not actually reach the value at these points. Thus, the function is discontinuous at \(x = n\pi\) for any integer value of \(n\).