If $f(x)=\left\{\begin{array}{cc}\frac{\sin \{\cos x\}}{x-\pi / 2}, & x \neq \frac{\pi}{2} \\ 1, & x=\pi / 2\end{array}\right.$, where {.} represents the fractional part function, then f(x), is

$\lim\limits _{x \rightarrow \pi / 2} f(x)$ exists, but f(x) is not continuous at $x=\pi / 2$

We have,

$\lim\limits_{x \rightarrow \pi / 2^{-}} f(x)=\lim\limits_{h \rightarrow 0} f(\pi / 2-h)$

$\Rightarrow \lim\limits_{x \rightarrow \pi / 2^{-}} f(x)=\lim\limits_{h \rightarrow 0} \frac{\sin \{\cos (\pi / 2-h)\}}{-h}$

$\Rightarrow \lim\limits_{x \rightarrow \pi / 2^{-}} f(x)=\lim\limits_{h \rightarrow 0} \frac{\sin \{\sin h\}}{-h}$

$\Rightarrow \lim\limits_{x \rightarrow \pi / 2^{-}} f(x)=\lim\limits_{h \rightarrow 0} \frac{\sin (\sin h)}{-h}$

$\Rightarrow \lim\limits_{x \rightarrow \pi / 2^{-}} f(x)=-\lim\limits_{h \rightarrow 0} \frac{\sin (\sin h)}{\sin h} \times \frac{\sin h}{h}=-1$

and,

$\lim\limits_{x \rightarrow \pi / 2^{+}} f(x)=\lim\limits_{h \rightarrow 0} f(\pi / 2+h)$

$\Rightarrow \lim\limits_{x \rightarrow \pi / 2^{+}} f(x)=\lim\limits_{h \rightarrow 0} \frac{\sin \{\cos (\pi / 2+h)\}}{\frac{\pi}{2}+h-\frac{\pi}{2}}$

$\Rightarrow \lim\limits_{x \rightarrow \pi / 2^{+}} f(x)=\lim\limits_{h \rightarrow 0} \frac{\sin \{-\sin h\}}{h}$

$\Rightarrow \lim\limits_{x \rightarrow \pi / 2^{+}} f(x)=-\lim\limits_{h \rightarrow 0} \frac{\sin \{\sin h\}}{h}$

$\Rightarrow \lim\limits_{x \rightarrow \pi / 2^{+}} f(x)=-\lim\limits_{h \rightarrow 0} \frac{\sin (\sin h)}{\sin h} \times \frac{\sin h}{h}=-1$

Clearly, $\lim\limits_{x \rightarrow \pi / 2^{-}} f(x)=\lim\limits_{x \rightarrow \pi / 2^{+}} f(x) \neq f\left(\frac{\pi}{2}\right)$

So, $\lim\limits_{x \rightarrow \pi / 2} f(x)$ exists, but f(x) is not continuous at $x=\pi / 2$.