Finding $\lim\limits_{x\to 0} \frac{\lfloor x \rfloor}{x}$ and the different definitions of fractional part function. 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.
 A: 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.
A: 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 $\lim_{x \to 0} 1 - \frac{\{x\}}{x}$.
