The set of values of $$a$$ for which $$\lim_{x \to a} ([x-5] - [2x+2]) = 0$$, where $$[\zeta]$$ denotes the greatest integer less than or equal to $$\zeta$$ is equal to

$$[x - 5] = [x] - 5$$

$$[2x + 2] = [2x] + 2$$

$$f(x) = ([x] - 5) - ([2x] + 2)$$

$$f(x) = [x] - [2x] - 7$$

For this limit to exist and equal 0, the function must be exactly 0 in an open interval around $$a$$.

So, we need to solve the equation: $$[x] - [2x] - 7 = 0$$ 

$$[2x] - [x] = -7$$

$$x = n + f$$ ($$x$$ is being written as the combination of integral and fractional part)

$$[2(n + f)] - n = -7$$

$$[2n + 2f] - n = -7$$

$$(2n + [2f]) - n = -7$$ (Since $$2n$$ is an integer)

$$n + [2f] = -7$$

Since $$0 \le f < 1$$, the value of $$2f$$ lies in the range $$0 \le 2f < 2$$. Thus, $$[2f]$$ can only be either $$0$$ or $$1$$

Case 1: ($$[2f] = 0$$): $$0 \le f < 0.5 \implies n = -7 \implies x \in [-7, -6.5)$$

Case 2: ($$[2f] = 1$$): $$0.5 \le f < 1 \implies n = -8 \implies x \in [-7.5, -7)$$

Combining these, $$f(x) = 0$$ on the interval $$[-7.5, -6.5)$$.

For the two-sided limit $$\lim_{x \to a} f(x) = 0$$ to exist, $$f(x)$$ must be $$0$$ on both sides of $$a$$. At the endpoints ($$-7.5$$ and $$-6.5$$), the function jumps to a non-zero value, meaning the limit does not exist there. Therefore, $$a$$ must lie strictly inside the open interval.