Limit of floor function when $x$ goes infinity Is it true that $\lim_{x \to \infty} (\left \lfloor{x}\right \rfloor -x) = 0$, or alternatively, $\lim_{x \to \infty} \left \lfloor{x}\right \rfloor=x$? If so, how can we prove it using $\varepsilon$-$\delta$ method?
 A: To say that $\lim_{x \to \infty} ( \lfloor x \rfloor -x) = \text{something}$ does not at all imply that $\lim_{x \to \infty} \lfloor x \rfloor  = x+\text{something}$.  Any expression of the form $\left(\lim\limits_{w\to\text{something}} \text{something} \right)$, if it can be evaluated at all, must come to something not depending on the variable $w$ that is approaching something.  That variable is a bound variable.
The function $x\mapsto \lfloor x\rfloor - x$ is periodic with period $1$, i.e. every time $x$ increases by $1$ it starts over and repeats.  Such a function cannot have a limit at $\infty$ unless it is a constant function.
A: Hint: it it were true, then as the sequence 
$$x_n = n + \frac 12
$$is such as $x_n\to\infty$ then $\lfloor{x_n} - x_n\rfloor \to 0$. 
But
$$
\lfloor{x_n} - x_n\rfloor = n - \left(n+\frac12\right) = -\frac12
$$
A: Since f: x -> $\lfloor x \rfloor - x$ is not continuous, I'm not sure using a sequence as a counter example is the right thing to do. Need some proving at least
I would prove that f is 1-periodic and continuous on each interval of the forme ]k, k+1[ :
First you prove that : $\lfloor x+1 \rfloor  = \lfloor x \rfloor +1 $
Then you get : f(x+1 ) = $\lfloor x+1 \rfloor -(x+1) = \lfloor x \rfloor -x = f(x)   $
Then for x,y each element of ]k,k+[ : $\lfloor x \rfloor =\lfloor y \rfloor$
|f(x)-f(y)| =  |$\lfloor x \rfloor  -x -(\lfloor y \rfloor -y)| =  |x-y|$ -> f continuous on ]k,k+1[
So from these properties, f cannot have 0 as a limit when x-> ∞
