Calculating $\lim_{x \to 0}{\frac{ \lfloor x \rfloor}{\lfloor x \rfloor}}$ My question is about 
$$\lim_{x \to 0}{\frac{ \lfloor x \rfloor}{\lfloor x \rfloor}}$$
where the notation is the floor function. 
I've graphed it and it is 1 everywhere except for $[0, 1)$. 
So, I think the answer should be 1, according to the epsilon-delta definition, since the function isn't defined for $0^+$, we can only check at $0^-$.
Is this correct? 
Also, is it to correct to simply cancel both the numerator and denominator here?
Thanks.
 A: It depends on how you define limits. If you require that a function is defined on a whole punctured neighborhood of $x_0$ (where you take the limit at), then this function doesn't have a limit at $0$.
If your definition is

Let $f$ be defined on a set $D$ and let $x_0$ be an accumulation point of $D$ ($x_0$ is not required to belong to $D$). We say that
  $$
\lim_{x\to x_0} f(x)=l
$$
  if, for every $\varepsilon>0$, there exists $\delta>0$ such that
  $$\text{for all $x\in D$, if $0<|x-x_0|<\delta$, then $|f(x)-l|<\varepsilon$}$$

then you can state correctly that
$$
\lim_{x\to0}\frac{\lfloor x\rfloor}{\lfloor x\rfloor}=1
$$
Note 1. The condition that $x_0$ is an accumulation point of $D$ avoids problems with isolated points of the domain, so as to ensure uniqueness of the limit.
Note 2. With this definition, the limit from the right is the limit of the function $f$ restricted to the domain $D\cap(x_0,\infty)$. Similarly for the limit from the left.
A: Yes the function is defined only for $x<0$ and, according to the more general definition of limit, we have that
$$\lim_{x \to 0}{\frac{ \lfloor x \rfloor}{\lfloor x \rfloor}}=\lim_{x \to 0^-}{\frac{ \lfloor x \rfloor}{\lfloor x \rfloor}}=\lim_{x \to 0^-}1=1$$
For the different definition of limits refer also to the related


*

*Does this limit exist on $\mathbb R^2$
