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We have $x \in [c, \infty]$ with $g: [c, \infty] \to \mathbb{R}$ , $g(x) = \lfloor x \rfloor + \sqrt{x+\lfloor x \rfloor}$

The question is for which $c$ is the function $g$ continous?

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There is no such $c$. The function jumps by an amount of $1+\sqrt {1+n}$ at $x =n$ so it is discontinuous at $n$ for any $n >c$.