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?

• No, it's not true. And $\lim_{x\to\infty} \lfloor x\rfloor = x$ doesn't make sense. However, we have $\lim_{x\to\infty} \frac{\lfloor x\rfloor}{x} = 1$. – Daniel Fischer Oct 19 '14 at 23:52
• Take $x_n = n + \frac{1}{2}$ for example, then $x_n \to +\infty$ but $\lfloor{x_n} \rfloor -x_n = -\frac{1}{2}$ – Petite Etincelle Oct 19 '14 at 23:54
• I dont see how the two statements are equivalent. Also, I don't see how the limit is zero. – FormerMath Oct 19 '14 at 23:55
• I understand that it is not true. But my question actually came from solving a problem in probability theory, which my assumption was the floor(X) converges to X "almost surely." Can I argue the convergence if we are talking about "almost surely"?? – aaka Oct 20 '14 at 0:04

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.

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 $$\lfloor{x_n} - x_n\rfloor = n - \left(n+\frac12\right) = -\frac12$$

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-> ∞