$f(x)=\sqrt{|x|-E(x)}$ On pose la fonction
$$f(x)=\sqrt{|x|-E(x)}$$
Est-ce que $f$ admet une limite en $p$ avec $p \in Z^*$?
Does $f$ admit a limit in $p$?
J'ai essayé d'encadrer l'expression et de diviser en deux cas $p > 0$ et $p < 0$ 
avec $E(x)$ est la partie entiere de $x$.

Let $f$ be defined as
$$
f(x) = \sqrt{\lvert x\rvert - \lfloor x\rfloor}.
$$
Fix $p\in\mathbb{Z}\setminus\{0\}$. Does $f$ have a limit at $p$?
I tried to upper and lower bound the function, dividing into the cases $p>0$ and $p<0$.
 A: Take $p \in \mathbb{Z}$ with $p\neq 0$. 


*

*Case $p>0$. For any $h \in (0,1)$, $p+h>0$, $p-h > 0$, $\lfloor p+h\rfloor = p$ and $\lfloor p-h\rfloor=p-1$, so that $$f(p+h) = \sqrt{\lvert p+h\rvert - \lfloor p+h\rfloor} = \sqrt{p+h - p} = \sqrt{h}\xrightarrow[h\to0^+]{} 0$$
but
$$f(p-h) = \sqrt{\lvert p-h\rvert - \lfloor p-h\rfloor} = \sqrt{p-h - (p-1)} = \sqrt{1-h}\xrightarrow[h\to0^+]{} 1$$
so $f$ is not continuous at $p$: both $\lim_{p^+} f$ and $\lim_{p^-} f$ exist,  but $\lim_{p^+} f\neq \lim_{p^-} f.$ 

*Case $p<0$. For any $h \in (0,1)$, $p+h<0$, $p-h < 0$, $\lfloor p+h\rfloor = p$ and $\lfloor p-h\rfloor=p-1$, so that $$f(p+h) = \sqrt{\lvert p+h\rvert - \lfloor p+h\rfloor} = \sqrt{-p-h - p} = \sqrt{-2p-h}\xrightarrow[h\to0^+]{} \sqrt{-2p}$$
but
$$f(p-h) = \sqrt{\lvert p-h\rvert - \lfloor p-h\rfloor} = \sqrt{-p+h - (p-1)} \xrightarrow[h\to0^+]{} \sqrt{-2p+1}$$
so $f$ is not continuous at $p$.

Note that you can visualize the discontinuities by looking at the graph of the function.
