Let a be an integer such that $\mathop {\lim }\limits_{x \to 7} {{18 - [1 - x]} \over {[x - 3a]}}$ exists, where [t] is greatest integer $\le$ t. Then a is equal to :

<p>$\mathop {\lim }\limits_{x \to 7} {{18 - [1 - x]} \over {[x - 3a]}}$ exist & $a \in I$.</p> <p>$= \mathop {\lim }\limits_{x \to 7} {{17 - [ - x]} \over {[x] - 3a}}$ exist</p> <p>$$RHL = \mathop {\lim }\limits_{x \to {7^ + }} {{17 - [ - x]} \over {[x] - 3a}} = {{25} \over {7 - 3a}}$$ $\left[ {a \ne {7 \over 3}} \right]$</p> <p>$$LHL = \mathop {\lim }\limits_{x \to {7^ - }} {{17 - [ - x]} \over {[x] - 3a}} = {{24} \over {6 - 3a}}$$ $\left[ {a \ne 2} \right]$</p> <p>For limit to exist</p> <p>$LHL = RHL$</p> <p>${{25} \over {7 - 3a}} = {{24} \over {6 - 3a}}$</p> <p>$\Rightarrow {{25} \over {7 - 3a}} = {8 \over {2 - a}}$</p> <p>$\therefore$ $a = - 6$</p>