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I understand that $\lim\limits_{x\to 0} \frac{\lfloor x \rfloor}{x}$ does not exist because RHL is $0$ and LHL is $\infty$. However, when I tried to calculate the limit of the equivalent expression $1-\frac{\{x\}}x$ I discovered that there were three distinct ways to define $\{x\}$. Which one of these definitions, if any, gives us the same LHL and RHL as in the initial case and shouldn't that definition alone be accepted? Related.

rtybase
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ZSMJ
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  • What three definitions of ${x}$ are you referring to? – coffeemath Feb 12 '19 at 11:19
  • @coffeemath They are mentioned here https://math.stackexchange.com/questions/1461955/what-is-operatornamefracx-or-x and taken from the wikipedia article. – ZSMJ Feb 12 '19 at 11:23
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    There is only one way to define ${,x,}$, namely, it is $x-\lfloor x\rfloor$. – Gerry Myerson Feb 12 '19 at 12:12
  • @GerryMyerson Could you have a look at this article https://en.wikipedia.org/wiki/Fractional_part#For_negative_numbers and confirm that the two other ways to define ${x}$ for negative $x$ namely, $|x|-\lfloor |x| \rfloor$ and $x-\lceil x \rceil$ are incorrect. – ZSMJ Feb 12 '19 at 13:46
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    I don't think Gerry can confirm those other definitions are incorrect. They are different definitions, the article itself states <<let it be $−1.3$, its fractional part will be $0.7$ according to the first definition, $0.3$ according to the second definition, and $−0.3$ according to the third definition>>. However, you, in your question, are using only one of those definitions, precisely $x=\lfloor x\rfloor +{x}$. I.e. the most familiar one. – rtybase Feb 12 '19 at 14:37
  • Hopefully this article will clarify the confusion. At <<perhaps most other mathematicians, use the different definition ... (right figure)>>. – rtybase Feb 12 '19 at 14:48
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    @rtybase Thank you for your patience and your help. – ZSMJ Feb 12 '19 at 14:50
  • Don't believe everything you read on Wikipedia, Rhaldryn. – Gerry Myerson Feb 12 '19 at 20:51

2 Answers2

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Think about the definition of the limit and what happens when

  • $\color{red}{x\rightarrow 0^{-}}$ then $\lfloor x \rfloor=-1$ for very small $\color{red}{-1<x<}0 \Rightarrow \frac{\lfloor x \rfloor}{x}=\frac{-1}{x}=\frac{1}{-x} \rightarrow \color{red}{+\infty}$
  • $\color{blue}{x\rightarrow 0^{+}}$ then $\lfloor x \rfloor=0$ for very small $\color{blue}{0<x<}1 \Rightarrow \frac{\lfloor x \rfloor}{x}=0 \rightarrow \color{blue}{0}$

Now, if you look at $\{x\}\overset{def}{=}x-\lfloor x \rfloor \Rightarrow 0\leq \{x\}<1$ $$\frac{\lfloor x \rfloor}{x}=1-\frac{\{x\}}{x}$$ you still have 2 different values

  • $\color{red}{x\rightarrow 0^{-}}$ then for very small $\color{red}{-1<x<0} \Rightarrow 1-\frac{\{x\}}{x}=1+\frac{\{x\}}{-x}>\color{red}{1}$. It is not too difficult to see from $x=\lfloor x \rfloor+\{x\}$ that we have $x=-1+\{x\} \Rightarrow 0=\lim\limits_{x\rightarrow0^{-}} x=-1+\lim\limits_{x\rightarrow0^{-}}\{x\}$. Or $\lim\limits_{x\rightarrow0^{-}}\{x\}=1$. Finally $$1-\frac{\{x\}}{x}=1+\frac{\{x\}}{-x}>1+\frac{1-\varepsilon}{-x}\rightarrow \color{red}{+\infty}$$
  • $\color{blue}{x\rightarrow 0^{+}}$ then for very small $\color{blue}{0<x<1} \Rightarrow 1-\frac{\{x\}}{x}=1-\frac{\{x\}}{\{x\}}=\color{blue}{0}$

In both cases the limit doesn't exist.

rtybase
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  • Is $lim_{x \to 0^{-}} 1- \frac{{x}}{x}=+ \infty$ too? – ZSMJ Feb 12 '19 at 13:59
  • Yes it is. Because ${x}$ will approach $1$. I didn't want to complicate the things, since it is clear that the limit doesn't exist. – rtybase Feb 12 '19 at 14:20
  • Why does this graph suggest that limit from both sides is $0$ https://www.wolframalpha.com/input/?i=1-frac(x)%2Fx – ZSMJ Feb 12 '19 at 14:37
  • Because you are using a different function, not the ${x}=x-\lfloor x \rfloor$ one. – rtybase Feb 12 '19 at 14:40
  • Using the most familiar definition $x= \lfloor x \rfloor + {x}$, I'm having trouble proving $lim_{x \to 0^{-}} 1-\frac{{x}}{x}=+\infty$ rigorously. – ZSMJ Feb 12 '19 at 14:48
  • @Rhaldryn I updated my answer. Let me know if it's more clear now. – rtybase Feb 12 '19 at 15:00
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    @I think I get it now, thanks a lot. – ZSMJ Feb 12 '19 at 15:28
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In general, if $f(x) = g(x)$ for all $x$ except, possibly, for $0$, then $\lim_{x \to 0} f(x) =\lim_{x \to 0} g(x)$ (even in the strong sense that the one exists if and only the other exists, and even if the limit happens to be $\pm \infty$). This is rather obvious, as the definition of $\lim$ doesn't care about the expressions you use to write $f$ and $g$ down and doesn't care about the value $x = 0$. The same holds for $\lim_{x \to 0^+}$ and $\lim_{x \to 0^-}$.

So, even without looking the three distinct ways to define $\{x\}$, if for those definitions $\frac{\lfloor x\rfloor}{x}$ is really equal to $1 - \frac{\{x\}}{x}$ (for all $x \neq 0$), then they all give the same left-hand- and right-hand-limit.

Now, the Wikipedia page gives three distinct ways to define $\{x\}$ for negative numbers, but only for one of them do you get $\frac{\lfloor x\rfloor}{x} = 1 - \frac{\{x\}}{x}$ (for all $ x \neq 0$), namely $\{x\} = x - \lfloor x \rfloor$. So that's the only one that's relevant if you're comparing $\lim_{x \to 0} \frac{\lfloor x\rfloor}{x}$ to $\lim_{x \to 0} 1 - \frac{\{x\}}{x}$.

Magdiragdag
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