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?

  • $\begingroup$ For the first limit, think of $x$ as $1.99$. Then $\lfloor x\rfloor=1$ and $\lfloor -x\rfloor=-2$. For the second limit, think of $x$ as $2.01$. Then, $\lfloor x\rfloor = 2$ and $\lfloor -x\rfloor = -3$. Note that the double bracket notation is frequently used for the integral part of a function instead of the greatest integer function (so make sure to double check your notation). $\endgroup$ – Michael Burr Feb 16 '18 at 10:07
  • $\begingroup$ Thank you (just stared Calculus I, trying to get-the-hang-of things) $\endgroup$ – bb411 Feb 16 '18 at 10:19

From the left, it should be $$(1)+(-2) = -1$$ and from the right, it should be $$(2)+(-3) = -1$$ so the limit is $-1$.


  • 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 ${\displaystyle{\lim_{x \to p}}}f(x) = -1$.


The floor enjoys an integer translation property:

$$\lfloor x+n\rfloor=\lfloor x\rfloor+n.$$


$$\lfloor \epsilon+2\rfloor+\lfloor-\epsilon-2\rfloor=\lfloor \epsilon\rfloor+\lfloor-\epsilon\rfloor+2-2=\lfloor \epsilon\rfloor+\lfloor-\epsilon\rfloor$$ is independent of the sign of $\epsilon$ and the limit exists.


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 $\mathbb Z$.


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