JEE Main · 2023 · 24 Jan 2023 · Shift 1 · Limits

Since $5$ and $2$ are integers, $[x-5]=[x]-5$ and $[2x+2]=[2x]+2$, so the expression is $[x]-[2x]-7$, and the limit is $0$ exactly when $[x]-[2x]\to 7$ from both sides of $a$. Writing $x=n+f$ with $n$ an integer and $0\le f<1$, one gets $[x]-[2x]=-n$ for $0\le f<\frac{1}{2}$ and $-n-1$ for $\frac{1}{2}\le f<1$. This equals $7$ on $[-7,-6.5)$ (with $n=-7$) and on $[-7.5,-7)$ (with $n=-8$), and the two pieces agree in value at $x=-7$. Hence the step function stays at $7$ throughout $[-7.5,-6.5)$, so a genuine two-sided limit of $7$ exists at every interior point, while at $x=-7.5$ and $x=-6.5$ the one-sided values differ. The set of admissible $a$ is $(-7.5,-6.5)$.