Question

If $f\left(x\right)=\underset{n\to \infty }{lim}=\underset{n\to \infty }{lim}(\sin x{)}^{2n}$, then f is

• A. continuous at $x=\pi$
• B. discontinuous at $x=\pi /2$
• C. discontinuous at $x=-\pi /2$
• D. discontinuous at an infinite number of points.

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A continuous at $x=\pi$
B discontinuous at $x=\pi /2$
C discontinuous at $x=-\pi /2$
D discontinuous at an infinite number of points.

$\underset{n\to \infty }{lim}{x}^{2n}\{\begin{array}{cc}0& \text{if}\left|x\right|<1\\ 1& \text{if}\left|x\right|=1\end{array}$
$\Rightarrow f\left(x\right)=\underset{n\to \infty }{lim}(\sin x{)}^{2n}=\{\begin{array}{cc}0& \text{if}\left|\sin x\right|<1\\ 1& \text{if}\left|\sin x\right|=1\end{array}$
This shows that f is continuous for all x, except possibly when $\left|\sin x\right|=1$,
i.e., when $x=(2k+1)\pi /2(k\in I)$. For these points, we have
$\underset{x\to (2k+1)\frac{\pi }{2}}{lim}f\left(x\right)=0\ne 1=f\left(\right(2k+1\left)\frac{\pi }{2}\right)$
Hence f(x) is discontinuous at these points.