Does this limit exist for integers'part?

Question

I m studying Calculus. We are focused on floor and ceiling function. The floor of a real number $x$, is defined to be the largest integer no larger than $x$. The ceiling of a real number $x$, is defined to be the smallest integer no smaller than $x$. Let $$f (x) = x- [| x |].$$ What values of $$a$$ exist? If we consider $$\lim_ {x\rightarrow a} f(x)$$ I need to prove it for the integers. I tried by myself: The limit of$$ f (x)$$ as $x$ approaches$$ a $$from left and right side. If we consider $$\lim_ {x\rightarrow a+} x- \bigl[| x |\bigr] =\lim_ {x\rightarrow a-} x-\bigl[| x |\bigr]$$ So, it must be equal. Any suggestions?. Thanks in advance.

Answer 1

I assume that by $[|x|]$ you mean the floor function, more commonly denoted by $\lfloor x\rfloor$.

Both floor and ceil functions are continuous on any interval of the form $(n,n+1)$ for $n\in\mathbb{Z}$. In fact they are constant there. And so your $f$ is continuous on them as well. Meaning the limit exists for any $a\in\mathbb{R}\backslash\mathbb{Z}$.

Now for $a\in\mathbb{Z}$ note that if $\epsilon>0$ is sufficiently small (i.e. $\epsilon<1$) then $f(a-\epsilon)=1-\epsilon$ while $f(a+\epsilon)=\epsilon$. And so $\lim_{x\to a^-}f(x)=1$ while $\lim_{x\to a^+}f(x)=0$.