[tex]f(x) =\lim\limits_{n\to \infty}\sin^{2n}x = \begin{cases} 0,\,\,|\sin x| < 1\\ 1,\,\, |\sin x| = 1 \end{cases} \\ \\\\ \Rightarrow f(x)=\begin{cases} 0,\,\,x\in [-\pi, \pi]\,\backslash\left\{-\frac{\pi}{2},\frac{\pi}{2}\right\}\\ 1,\,\, x\in \left\{-\frac{\pi}{2},\frac{\pi}{2}\right\}\end{cases}\\\\[/tex]
[tex]\textbf{Pentru }x = -\frac{\pi}{2}:\\ \\\lim\limits_{x\searrow -\frac{\pi}{2}}f(x) = 0,\,\,\lim\limits_{x\nearrow -\frac{\pi}{2}}f(x) = 0, \,\, f\left(-\frac{\pi}{2}\right) = 1\\ \\ \Rightarrow \text{punct de discontinuitate (cele 3 limite nu sunt egale).}\\ \\ \textbf{Pentru }x = \frac{\pi}{2}:\\ \\\lim\limits_{x\searrow \frac{\pi}{2}}f(x) = 0,\,\,\lim\limits_{x\nearrow \frac{\pi}{2}}f(x) = 0, \,\, f\left(\frac{\pi}{2}\right) = 1\\ \\ \Rightarrow \text{punct de discontinuitate (cele 3 limite nu sunt egale).}\\\\[/tex]
Răspuns:
c) k = 2
