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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 $p<0$.

Clement C.
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user315918
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1 Answers1

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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.

Clement C.
  • 67,323