What is the difference between f(x) =[|x|] (floor function) and f(x)=x-[|x|] (fractional part)?

Question

I'm studying Calculus. My teacher gave us a couple of exercises: first, prove $$\lim_{x \rightarrow a}\,[|x|]$$ (floor function) exists for values of $a$, and second, prove $$\lim_{x \rightarrow a}\,(x-[|x|])$$ (fractional part) exists for values of $a$. I know the definition for limits, I mean, the right and left sides must be equal. But, how do you find the interval for each one? How do you prove each limit for values of $a$? I think that it doesn't exist at all.

I assume that by $[|x|]$ you mean the floor function, more commonly denoted by $\lfloor x\rfloor$.

Both floor and ceiling functions are continuous on any interval of the form $(n,n+1)$ for $n\in\mathbb{Z}$. In fact they are constant there. And so your $f$ is continuous on them as well. Meaning the limit exists for any $a\not\in\mathbb{Z}$.

Now for $a\in\mathbb{Z}$ note that if $\epsilon>0$ is sufficiently small (i.e. $\epsilon<1$) then $f(a-\epsilon)=1-\epsilon$ while $f(a+\epsilon)=\epsilon$. And so $\lim_{x\to a^-}f(x)=1$ while $\lim_{x\to a^+}f(x)=0$. Is it correct?, how do you prove the other case?

Answer 1

Note that the floor function is continuous for $x \notin \mathbb{Z}$.

If $n \in \mathbb{Z}$ then $\lim_{x \uparrow n} \lfloor x \rfloor = n-1$ and $\lim_{x \downarrow n} \lfloor x \rfloor = n$, hence the floor function is not continuous at $n$.

Since the function $x \mapsto x$ is continuous, it follows that $g(x)= x- \lfloor x \rfloor$ shares the same points of continuity and discontinuity as the floor function.

To elaborate, if $x \in (n,n+1)$ for some $n \in \mathbb{Z}$ then $g$ is continuous and $g(x) = x-n$. Hence, $\lim_{x \uparrow n} g(x) = 1$, $\lim_{x \downarrow n} g(x) = 0$, so $g$ is not continuous at the integers.

Answer 2

Let $f(x) =[|x|] $. If $a\ne 0$ is an integer then, $\lim_{x\to a^{-}}f(x)=a-1$ and $\lim_{x\to a^{+}}=a. $ Hence, $\lim_{x\to a^{-}}f(x)\ne \lim_{x\to a^{+}} f(x) $.
Notice that for all $x\ne0$, $0\le f(x) \le |x|$ By squeeze theorem, $f$ has limit 0 at x=0.
Now show using epsilon delta definition of limit that if $r$ is a non integer point then $\lim_{x\to r}f(x) =[r] $ Can you take it from here?