The limit of the sum of two floor functions The question asks to calculate the limits where they exist, then the following limit is given:
$$\lim_{x\rightarrow2}(\lfloor x\rfloor +\left\lfloor -x\right\rfloor),   \text{where}\lfloor x\rfloor \text{is the floor function.} $$
Thus I approach it from the left hand side and then from the right hand side:
$$\lim_{x\rightarrow2^-}(\lfloor x\rfloor +\left\lfloor -x\right\rfloor)\textbf{ and} \lim_{x\rightarrow2^+}(\lfloor x\rfloor +\left\lfloor -x\right\rfloor) \\=(1)+(-3)\phantom{help me!} =(2)+(-2)$$
Am I correct in making the following conclusion:
Since $$\text{Since} \lim_{x\rightarrow2^-}(\lfloor x\rfloor +\left\lfloor -x\right\rfloor)\neq\lim_{x\rightarrow2^+}(\lfloor x\rfloor +\left\lfloor -x\right\rfloor),\text{ we have that}\\ \lim_{x\rightarrow2}(\lfloor x\rfloor +\left\lfloor -x\right\rfloor) \text{ does not exist.}
$$
or are my calculations wrong?
 A: From the left, it should be
$$(1)+(-2) = -1$$
and from the right, it should be
$$(2)+(-3) = -1$$
so the limit is $-1$.

Explanation:


*

*If $x$ is a little less than $2$, strictly between $1$ and $2$, then $-x$ is strictly between $-2$ and $-1$.

*If $x$ is a little more than $2$, strictly between $2$ and $3$, then $-x$ is strictly between $-3$ and $-2$.


More generally, the behavior of the function $f(x) = \lfloor{x}\rfloor + \lfloor{-x}\rfloor$ can be analyzed as follows . . .


*

*If $x$ is an integer, then $\lfloor{x}\rfloor=x$, and $\lfloor{-x}\rfloor=-x$, hence 
$$f(x) = \lfloor{x}\rfloor + \lfloor{-x}\rfloor = (x) + (-x) = 0$$

*If $x$ is not an integer, then $a < x < a+1$ for some integer $a$, hence $-a-1 < -x < -a.\;$Then $\lfloor{x}\rfloor=a$, and $\lfloor{-x}\rfloor=-a-1$, which yields
$$f(x) = \lfloor{x}\rfloor + \lfloor{-x}\rfloor = (a) + (-a-1) = -1$$


It follows that for all $p\in \mathbb{R}$, we have  ${\displaystyle{\lim_{x \to p}}}f(x) = -1$.
A: The floor enjoys an integer translation property:
$$\lfloor x+n\rfloor=\lfloor x\rfloor+n.$$
Then
$$\lfloor \epsilon+2\rfloor+\lfloor-\epsilon-2\rfloor=\lfloor \epsilon\rfloor+\lfloor-\epsilon\rfloor+2-2=\lfloor \epsilon\rfloor+\lfloor-\epsilon\rfloor$$ is  independent of the sign of $\epsilon$ and the limit exists.
A: If $r\in(0,1)$ then 


*

*$\lfloor 2-r\rfloor+\lfloor -2+r\rfloor=1+(-2)=-1$

*$\lfloor 2+r\rfloor+\lfloor -2-r\rfloor=2+(-3)=-1$


So there is a flaw in your calculations.
The limit exists and equals $-1$.
Also it is handsome to note that $\lfloor -x\rfloor=-\lceil x\rceil$ so that $$\lfloor x\rfloor+\lfloor -x\rfloor=\lfloor x\rfloor-\lceil x\rceil$$
So the function takes value $-1$ on $\mathbb R\setminus\mathbb Z$ and takes value $0$ on $\mathbb Z$.
