Hint: The function $g(x)=3x-\frac12$ is increasing, and hence, for $\frac16\le x<\frac12$, we have $g\left(\frac16\right)\le g(x)<g\left(\frac12\right)$.

Since we are interested in the behavior of $\left\lfloor 3x-\frac{1}{2}\right\rfloor$ as $x$ tends to $\frac{1}{2}$ from the left, we may assume that $\frac{1}{6}<x<\frac{1}{2}$. Then

$$\frac{3}{6}<3x<\frac{3}{2}$$

so

$$\frac{3}{6}-\frac{1}{2}<3x-\frac{1}{2}<\frac{3}{2}-\frac{1}{2}$$

which reduces to $0<3x-\frac{1}{2}<1$. It follows that $\left\lfloor 3x-\frac{1}{2}\right\rfloor=0$, so

$$\lim_{x\to\frac{1}{2}^-}\left\lfloor 3x-\frac{1}{2}\right\rfloor=0$$