Therefore function is defined as $f(x) = \left\{ \begin{array}{l}\frac{{\sin \,[x]}}{{[x]}}\,\,( - 1 \le x < 0)\\\;\;\;\;\;0\;\;(0 \le x < 1)\end{array} \right.$
$\therefore$ Left hand limit $ = \mathop {\lim }\limits_{x \to 0 - } f(x) = \mathop {\lim }\limits_{x \to 0 - } \,\frac{{\sin \,[x]}}{{[x]}}$
$ = \frac{{\sin \,( - 1)}}{{ - 1}} = \sin {1^c}$
Right hand limit $= 0$.
Hence limit doesn't exist.
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Match each entry in $List-I$ to the correct entry in $List-II$.
| $List-I$ | $List-II$ |
| ($P$) $\alpha$ equals | ($1$) $(-2,4)$ |
| ($Q$) $r$ equals | ($2$) $\sqrt{5}$ |
| ($R$) $A_1$ equals | ($3$) $(-2,6)$ |
| ($S$) $B_1$ equals | ($4$) $5$ |
| ($5$) $(2,4)$ |
The correct option is