Since , for we have The limit of the left and right expressions as goes to infinity is , so the same holds for the middle, by the squeeze theorem.

Hint: In this problem, we have to know about the limit of the greatest integer function. We should know that the greatest integer is also known as floor function, which is written as \\[f\\left( x \\right)=\\left\\lfloor x \\right\\rfloor \\]. The value of \\[\\left\\lfloor x \\right\\rfloor \\] is the largest integer that is less than or equal to x. here we can see about the limits.Complete step-by-step answer:We know that the greatest integer function is the value of \\[\\left\\lfloor x \\right\\rfloor \\] which is the largest integer that is less than or equal to x.We have the following limits for the greatest integer function, they are\\[\\begin{align}  & \\Rightarrow \\displaystyle \\lim_{x \\to +\\infty }\\left\\lfloor x \\right\\rfloor =+\\infty \\\\  & \\Rightarrow \\displaystyle \\lim_{x \\to -\\infty }\\left\\lfloor x \\right\\rfloor =-\\infty \\\\ \\end{align}\\] Where if n is any integer (positive or negative), then we have \\[\\begin{align}  & \\Rightarrow \\displaystyle \\lim_{x \\to +{{n}^{-}}}\\left\\lfloor x \\right\\rfloor =n-1 \\\\  & \\Rightarrow \\displaystyle \\lim_{x \\to {{n}^{+}}}\\left\\lfloor x \\right\\rfloor =n \\\\ \\end{align}\\] Here we can see that the left and right limits differ at any integer and the function is discontinuous.We should know that for a real number which is not an integer, then we will have the limits as\\[\\Rightarrow \\displaystyle \\lim_{x \\to a}\\left\\lfloor x \\right\\rfloor =\\left\\lfloor a \\right\\rfloor \\] Hence the left and right limits agree at any other real number and the function will be continuous.Note: We should first understand the concept of limits and greatest integer function. We should remember that the value of \\[\\left\\lfloor x \\right\\rfloor \\] is the largest integer that is less than or equal to x. We should also note that for a real number which is not an integer we will have a different limit for which the function will be continuous.

Try putting $x=0^+$ (meaning a bit larger than zero) and $x=0^-$ (a bit smaller than zero) and see what happens.

in the first case, $-1<x-1<0$ and so $[x-1]=-1$ and with a similair argument, $[1-x]=0$

And so in the first example, if $x$ is a bit larger than $0$, the result is $0$.

Do the same thing when $x=0^-$ to see if it is the same answer. if it is, then the answer is $0$. if it isn't, then there is no limit.

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

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