# Limit of $f(x) = \lfloor x \rfloor + \sqrt{x - \lfloor x \rfloor}$

I'm doing Spivak's Calculus book and one of the exercises from 5th chapter says

4. For each of the functions [...], decide for which numbers a the limit $lim_{x \to a}\ f(x)$ exists.

i) $f(x) = \lfloor x \rfloor$

ii) $f(x) = x - \lfloor x \rfloor$

iii) $f(x) = \sqrt{x - \lfloor x \rfloor}$

iv) $f(x) = \lfloor x \rfloor+ \sqrt{x - \lfloor x \rfloor}$

$...$

I found that, for $i$, $ii$ and $iii$, the limit doesn't exist when $a = \lfloor a \rfloor$, namely, when a is an integer. But since in iv I have the floor function being used twice (once incrementing once decrementing), I'm not sure if I'm supposed to use the $x-1 \lt \lfloor x \rfloor \le x$ in the same way. What am I supposed to do here? How should I find inequalities for the new function?

• Please don't change the wording of the problem as seen in the book, @MITjanitor. Thanks. – FRD Mar 26 '13 at 18:55

## 2 Answers

Well, yes, you can use it, and it yields $$0=x-x\le x-\lfloor x\rfloor < x-(x-1)=1$$ thus, $0\le\sqrt{x-\lfloor x\rfloor}<1$ (it is just the square function defined on $[0,1)$ and made periodic with period $1$). This function ($x\mapsto \sqrt{x-\lfloor x\rfloor}$) is therefore continuous on $\Bbb R\setminus\Bbb Z$, and so is $x\mapsto \lfloor x\rfloor$.

You still have to check, perhaps the left and right limits at integers became the same if we add these.

Let $n\in\Bbb Z$, then, if $x\to n$ from below, then as $\lfloor x\rfloor=n-1$ for $x<n$ near, we have $$\lim_{x\to n^-}\lfloor x\rfloor=n-1$$ And similarly, $\lim_{x\to n^+}\lfloor x\rfloor=n$. (This in itself shows why $\lim_{x\to n}\lfloor x\rfloor$ doesn't exist.) For the other one we have $$\lim_{x\to n^-} \sqrt{x-\lfloor x\rfloor}=1 \\ \lim_{x\to n^+} \sqrt{x-\lfloor x\rfloor}=0$$ So, adding these: $$\lim_{x\to n^-} \lfloor x\rfloor+\sqrt{x-\lfloor x\rfloor}=n-1+1=n \\ \lim_{x\to n^+} \lfloor x\rfloor+\sqrt{x-\lfloor x\rfloor}=n+0=n \, .$$ See it also on WolframAlpha.

• but what about letter iv? It's $f(x) = \lfloor x \rfloor + \sqrt{x - \lfloor x \rfloor}$. – FRD Mar 26 '13 at 18:49
• But the book says the limit doesn't exist when a is integer in neither $\lfloor x \rfloor$ nor $\sqrt{x - \lfloor x \rfloor}$... – FRD Mar 26 '13 at 18:53
• I'm sorry to bother, but could you write a throughout explanation of "checking if the left and right limits at integers become the same"? – FRD Mar 26 '13 at 18:59

Note that for each $n\in\mathbb{Z}$ and $n\le x\lt n+1$, we have $$f(x)=\lfloor x\rfloor+\sqrt{x-\lfloor x\rfloor}=n+\sqrt{x-n\vphantom{\lfloor}}\tag{1}$$ Thus \begin{align} \color{#00A000}{\lim\limits_{x\to n^+}f(x)} &=n+\lim\limits_{x\to n^+}\sqrt{x-n\vphantom{\lfloor}}\\ &\color{#00A000}{=n}\tag{2}\\ \color{#C00000}{\lim\limits_{x\to n+1^-}f(x)} &=n+\lim\limits_{x\to n+1^-}\sqrt{x-n\vphantom{\lfloor}}\\ &\color{#C00000}{=n+1}\\[6pt] \color{#C00000}{\lim\limits_{x\to n^-}f(x)} &\color{#C00000}{=n}\tag{3} \end{align} Thus, $(2)$ and $(3)$ show that for $n\in\mathbb{Z}$, $$\lim_{x\to n}f(x)=n\tag{4}$$ Therefore, $f(x)$ is continuous at each $x\in\mathbb{Z}$.

For $x\in(n,n+1)$, $(1)$ shows that $f(x)$ is a composition of continuous functions in a neighborhood of $x$. Therefore, $f(x)$ is continuous for all $x\not\in\mathbb{Z}$.

Thus, $f(x)$ is continuous for all $x$.