Option : (C).
Formula : -
(i) \(\lim\limits_{x \to 0}\frac{log(1-x)}{x}\) = 1 and \(\lim\limits_{x \to 0}\frac{sin\,x}{x}\) = 1
(ii) A function f(x) is said to be continuous at a point x=a of its domain, if \(\lim\limits_{x \to a}f(x)\) = f(a)
\(\lim\limits_{x \to a^+}f(a+h)\) = \(\lim\limits_{x \to a^-}f(a-h)\) = f(a)
(iii) \(\lim\limits_{x \to a}{\{f(x).g(x)}\}\) = 1.m,
Where \(\lim\limits_{x \to a}f(x)\) = 1, \(\lim\limits_{x \to a}g(x)\) = m
Given :-
\(f(x) = \begin{cases} \frac{1-sinx}{(\pi-2x)^2}.\frac{log\,sin\,x}{(log(1+\pi^2-4\pi x+4x^2))},x ≠\frac{\pi}{2}\\k,x=\frac{\pi}{2} \end{cases} \)
Function f(x) is continuous at x = \(\frac{\pi}{2}\)
\(\lim\limits_{x \to \frac{\pi}{2}}f(x)\) = \(f(\frac{\pi}{2})\)
Using substitution method
If \(\frac{\pi}{2}\)- x, then
Using standard limit formula (i)
k = \(-\frac{1}{64}\)