If \[[ \text{  }]\] denotes the greatest integer function then \[f(x)=[x]+[x+\dfrac{1}{2}]\]
A. is continuous at \[x=\dfrac{1}{2}\]
B. is discontinuous at \[x=\dfrac{1}{2}\]
C. $\displaystyle \lim_{x \to \dfrac{1}{2}+}f(x)=2$
D. $\displaystyle \lim_{x \to \dfrac{1}{2}-}f(x)=1$

Hint: Possibility of discontinuity of greatest integer function \[[x+a]\] is only when \[x+a\] becomes integer , we can’t directly say that it is discontinuous at that point for that we have to check using continuity equation ,so in given question \[f(x)=[x]+[x+\dfrac{1}{2}]\], we first check continuity on \[x=\dfrac{1}{2}\]
By using equation
For continuity \[f(x)=f({{x}^{-}})=f({{x}^{+}})\]
So will put \[x=\dfrac{1}{2}\] and check accordingly which option satisfies

Complete step-by-step answer:
We are given a greatest integer function then \[f(x)=[x]+[x+\dfrac{1}{2}]\] , we have check for its continuity and for checking a functions continuity we have one formula that is \[f(x)=f({{x}^{-}})=f({{x}^{+}})\] and we know that possibility of discontinuity of greatest integer function
\[[x+a]\] is only when \[x+a\] becomes an integer, given in the question all option lies around \[x=\dfrac{1}{2}\]so we will check for \[x=\dfrac{1}{2}\] there can be discontinuity
Using greatest integer function property \[[x]=0\], when x is between \[[0,1)\] and \[[x]=1\] when x is between \[[1,2)\]
\[f(\dfrac{1}{2})=[\dfrac{1}{2}]+[\dfrac{1}{2}+\dfrac{1}{2}]=0+1=1\]
Similarly, on putting \[x={{\dfrac{1}{2}}^{+}}\]
\[f({{\dfrac{1}{2}}^{+}})=[{{\dfrac{1}{2}}^{+}}]+[{{\dfrac{1}{2}}^{+}}+\dfrac{1}{2}]=0+1=1\]
Similarly, on putting \[x={{\dfrac{1}{2}}^{-}}\]
\[f({{\dfrac{1}{2}}^{-}})=[{{\dfrac{1}{2}}^{-}}]+[{{\dfrac{1}{2}}^{-}}+\dfrac{1}{2}]=0+0=0\], this time in greatest integer function value becomes less then 1 result into 0
We can see very clearly that \[f(\dfrac{1}{2})=f({{\dfrac{1}{2}}^{-}})\ne f({{\dfrac{1}{2}}^{+}})\] , we confirms that \[f(x)=[x]+[x+\dfrac{1}{2}]\] is discontinuous at \[x=\dfrac{1}{2}\]
Now checking other option like (c) and (d)
In which we are asked to find the limit at \[x\to \dfrac{1}{2}\]
$$\displaystyle \lim_{x \to \dfrac{1}{2}+}f(x)=2$$ , for this put \[x={{\dfrac{1}{2}}^{+}}\] in given equation that we have already putted an got
\[f({{\dfrac{1}{2}}^{+}})=[{{\dfrac{1}{2}}^{+}}]+[{{\dfrac{1}{2}}^{+}}+\dfrac{1}{2}]=0+1=1\]
Similarly
$$\displaystyle \lim_{x \to \dfrac{1}{2}-}f(x)=1$$, for this put \[x={{\dfrac{1}{2}}^{-}}\] in given equation that we have already putted an got
\[f({{\dfrac{1}{2}}^{-}})=[{{\dfrac{1}{2}}^{-}}]+[{{\dfrac{1}{2}}^{-}}+\dfrac{1}{2}]=0+0=0\]
So, both options are wrong and

So, the correct answer is “Option (B)".

Note: Some of the student might think that for checking continuity of this \[f(x)=[x]+[x+\dfrac{1}{2}]\], they just check by equating greatest integer to 0 and getting value like here we get \[x=0\] and \[x=-\dfrac{1}{2}\] but we have to check it by equating greatest integer with all integer possible.