Continuity of $f(x)=[x]+ \sqrt{x-[x]}$ Consider the function $f(x)=[x]+ \sqrt{x-[x]}, \, x\in \Bbb R$ ; where "$[ \space ]$" denotes the greatest integer function. It is obvious that if $b$ is an integer, then 
$$\lim_{x\to b-} f(x)=\lim_{x\to b-}[x] + \sqrt { \lim_{x\to b-}x - \lim_{x\to b-}[x] } =b-1+\sqrt {b-(b-1)}=b-1+1=b$$ 
$$\lim_{x\to b+} f(x)=\lim_{x\to b+}[x] + \sqrt { \lim_{x\to b+}x - \lim_{x\to b+}[x] } =b+\sqrt {b-b}=b,$$ and 
$$f(b)=[b]+ \sqrt{b-[b]}=b+\sqrt {b-b}=b,$$ hence $f(x)$ is continuous for all integer values of $x$.  Is $f(x)$ continuous for all non-integer values of $x$ also, i.e. is it true that $f(x)$ is continuous for all $ x \in \Bbb R$ ? 
 A: If $x \notin \mathbb{Z}$, then for some $\epsilon >0$, $B(x,\epsilon) \cap \mathbb{Z} = \emptyset$. Then $\lfloor y \rfloor $ is constant for $y \in B(x,\epsilon)$. Let $k = \lfloor y \rfloor $. Note that if $y \in B(x,\epsilon)$, we have $y > k$.
For $y \in B(x,\epsilon)$, we have $f(y) = k + \sqrt{y-k}$, which is continuous (in fact smooth, since $y > k$).
$f$ is not differentiable at the integers.
A: Write $f(x)$ as
$$ f(x) = x + g(x) \quad \text{where} \quad g(x) = \sqrt{x - [x]} - (x - [x]). $$
It is clear that $g(x) = \sqrt{x} - x$ on $[0, 1)$ and $g(x+1) = g(x)$.
From this observation it is easy to prove that $g$ is continuous on $[0, 1]$ (the point $x = 1$ only matters), and then this continuity extends to $\Bbb{R}$ by periodicity of $g$. Therefore $g$ is continuous and hence the same is true for $f$.
A: Since continuity is a local property, for any $x_0$ not an integer, $f(x)$ is continuous at $x_0$ if and only if $g(x)=[x_0]+\sqrt{x-[x_0]}$ is continuous at $x_0$.  This is because $g(x)=f(x)$ in a neighborhood of $x_0$.  Since $g(x)$ is continuous at $x_0$, so is $f(x).$
