We need the set of all $a$ such that $\lim_{x\to a} \big([x-5]-[2x+2]\big)=0.$ Let $f(x)=[x-5]-[2x+2].$ We want points $a$ where the two-sided limit exists and equals $0$.

1. Rewrite using fractional-part idea

For any real $t$, $[t]=t-\{t\},$ where $\{t\}$ is the fractional part. So, \begin{align*} f(x)&=(x-5-{x-5})-(2x+2-{2x+2})\ &=-x-7+{2x+2}-{x-5}. \end{align*} This is not the most direct route for limits, so instead we analyze where the floor functions jump.

2. Where can the limit fail to be locally constant?

The function $[x-5]$ changes only when $x-5\in\mathbb Z$, i.e. $x\in \mathbb Z+5.$ The function $[2x+2]$ changes only when $2x+2\in\mathbb Z$, i.e. $x\in \frac{\mathbb Z-2}{2},$ which is the set of all integers and half-integers.

Hence $f(x)$ is constant on every open interval between consecutive half-integers. Therefore, if $a$ is not an integer or half-integer, then near $a$, $f(x)$ is constant, so $\lim_{x\to a} f(x)=f(a).$ Thus for such $a$, we only need $f(a)=0.$

3. Solve $[x-5]-[2x+2]=0$

We test $x$ in intervals of length $\frac12$.

Let $x=n+t,\qquad 0\le t<1.$ Then behavior depends on whether $t<\frac12$ or $t\ge \frac12$.

Case 1: $x=n+t$ with $0\le t<\frac12$

Then $2x+2=2n+2+2t$ has fractional part $2t<1$, so $[2x+2]=2n+2.$ Also, $[x-5]=[n+t-5]=n-5.$ Therefore, $f(x)=(n-5)-(2n+2)=-n-7.$ For this to be $0$, $-n-7=0 \implies n=-7.$ So in this case, $x\in[-7,-6.5).$

Case 2: $x=n+t$ with $\frac12\le t<1$

Now $2t\in[1,2)$, so $[2x+2]=2n+3.$ And still $[x-5]=n-5.$ Thus, $f(x)=(n-5)-(2n+3)=-n-8.$ For this to be $0$, $-n-8=0 \implies n=-8.$ So in this case, $x\in[-7.5,-7).$

Combining both cases, $f(x)=0 \iff x\in[-7.5,-7)\cup[-7,-6.5)=[-7.5,-6.5).$

4. Now impose the limit condition

We need $\lim_{x\to a} f(x)=0.$ Since $f$ is piecewise constant, this will hold exactly at points where a whole punctured neighborhood of $a$ lies inside the set where $f(x)=0$.

From above, $f(x)=0$ on $[-7.5,-6.5).$ But we must check endpoints and jump points carefully.

• For $a\in(-7.5,-7)$: neighborhood lies inside region where $f(x)=0$, so limit is $0$.
• For $a=-7$: on both sides, $f(x)=0$ (left interval $[-7.5,-7)$ and right interval $[-7,-6.5)$), and indeed $f(-7)=[-12]-[-12]=0.$ So limit is $0$.
• For $a\in(-7,-6.5)$: again neighborhood lies inside region where $f(x)=0$, so limit is $0$.

Now endpoints:

At $a=-7.5$

Take left and right limits:

• If $x\to -7.5^+$, then $f(x)=0$.
• If $x\to -7.5^-$, take for example $x\in(-8,-7.5)$. Then $n=-8$, $t<\frac12$, so $f(x)=-(-8)-7=1.$ Thus left and right limits differ, so limit is not $0$. Hence $a=-7.5$ is excluded.

At $a=-6.5$

• If $x\to -6.5^-$, then $f(x)=0$.
• If $x\to -6.5^+$, take $x\in(-6.5,-6)$. Then $n=-7$, $t\ge \frac12$, so $f(x)=-(-7)-8=-1.$ So the limit does not exist, hence not $0$. Thus $a=-6.5$ is excluded.

Therefore the required set is $(-7.5,-6.5).$

5. Match with options

This is option D.