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

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?

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.

• Ok. So Does it mean $$lim_{x \rightarrow a-} (x-[|x|]) \neq lim_{x \rightarrow a+} (x-[|x|])$$ ? Right? Apr 17, 2020 at 9:33
• @Charlie, it is correct if a is a non-zero integer.
– Koro
Apr 17, 2020 at 11:39
• Sure thanks. Just Im wondering what is the difference in the use of notation, up arrow and down arrow. But I found the answer right here math.stackexchange.com/questions/124489/limits-notation Apr 17, 2020 at 11:43

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?