The set of values of a for which _ x → a ( [x - 5] - [2x + 2]) = 0...

We have,

$\lim_{x \to a}\left(\left[x - 5\right] - \left[2x + 2\right]\right) = 0$

$⇒ \lim_{x \to a}\left(\left[x\right] - 5 - \left[2x\right] - 2\right) = 0$

$⇒ \lim_{x \to a}\left(\left[x\right] - \left[2x\right] - 7\right) = 0$

$⇒ \lim_{x \to a}\left(\left[x\right] - \left[2x\right]\right) = 7$

Now when $a \in I$ (integer)$then a - 2a = 7 ⇒ a = - 7$

Now when $a \in I + f$ where $f$ is fraction,

So, \left[I + f\right] - $\left[2I + 2f\right] = 7$

$⇒ I + \left[f\right] - 2I - \left[2f\right] = 7$

$⇒ I + 0 - 2I - \left[2f\right] = 7$

$⇒ - I - \left[2f\right] = 7$

Now taking case \left(1\right)$when$f \in \left(0, \dfrac{1}{2}\right)$</p><p>Then$⇒ - I - 0 = 7 ⇒ I = - 7 ⇒ a \in \left(- 7, - 6 . 5\right)$</p><p>Now taking case \left(2\right)$ when $f \in \left(\dfrac{1}{2}, 1\right)$

So, $- I - 1 = 7 ⇒ I = - 8 ⇒ a \in \left(- 7 . 5, 7\right)$

Hence, from both the cases we can say that $a \in \left(- 7 . 5, - 6 . 5\right)$.