# Let f(x)= $\lfloor x \rfloor+ \lfloor -x \rfloor$ Prove: The limit $\lim_{x \to x_0}f(x)$ exists for every $x_0\in R$.

Let $f:R \to R$ be $f(x)=\lfloor x \rfloor+ \lfloor -x \rfloor$ (floor function)

Prove or disprove: The limit $\lim_{x \to x_0}f(x)$ exists for every $x_0\in R$ and define what types of point discontinuities the function has, if any.

1.) if the limit exists we need to prove that both one-sided limits exist and are equal to each other.

$\lim_{x\to x_0^+}\lfloor x \rfloor+ \lfloor -x \rfloor$ = $\lfloor x_0^+ \rfloor+ \lfloor -x_0^+ \rfloor$= $x_0 - x_0 =0$

Similarly:

$\lim_{x\to {x_0-}}f(x)=0$

But this seems to only works for $x_0\gt0$

Picking $x_0=-4.3$ we get that f(x)=$\lfloor -4.3 \rfloor+ \lfloor 4.3 \rfloor=-5+4=-1$

Checking the one sided limits, we get that indeed for $x_0\lt 0$ exist and are equal to each other.

Does this make sense?

And for Discontinuities, it makes sense for them to be at 0 since picking something a little bit smaller than zero will give us -2, picking something a little bigger will give us 0. Therefore the discontinuity will be that the one-sided limits are not equal to each other (not sure of name.)

Would appreciate a quick look over to see if I'm right.

• write $x_0 = n + \epsilon$ where $n \in \mathbb{Z}, \epsilon \in [0,1)$, and consider the case $\epsilon > 0$ and $\epsilon = 0$ – reuns Jul 30 '16 at 14:38
• Where did you get $\lfloor x_0^+ \rfloor+ \lfloor -x_0^+ \rfloor=x_0-x_0$? This is just not true. For any non integer the value is $-1$ and for any integer the value is $0$. There are infinitely many points without a limit. – Elliot G Jul 30 '16 at 14:48
• @ElliotG ah thank you! So the limit doesn't exist since the one-sided limits are not equal to each other? – RonaldB Jul 30 '16 at 14:52
• Actually the limits equal one another but there is a third requirement that the function actually take on that value (which it doesn't). Hope that helps – Elliot G Jul 30 '16 at 14:57
• I have never seen any definition that wasn't equivalent to the delta-epsilon formation, in which case you need $f(x_0)$ to equal the left and right limits. – Elliot G Jul 30 '16 at 15:19

Simplify the expression of $f$: $$f(x)=\begin{cases} 0&\text{if } x\in\mathbf Z\\ -1 &\text{otherwise }\end{cases},$$ hence the limit is $-1$ at every point.
Let $x_0\in\mathbb{R}-\mathbb{Z}$ we have $$\lim_{x\to x_0^+}\lfloor x \rfloor=\lim_{x\to x_0^-}\lfloor x \rfloor=\lim_{x\to x_0}\lfloor x \rfloor=\lfloor x_0 \rfloor$$ so $$\lim_{x\to x_0}f(x)=\lim_{x\to x_0}\lfloor x \rfloor+\lfloor -x \rfloor=\lfloor x_0 \rfloor+\lfloor -x_0 \rfloor=\lfloor x_0 \rfloor-\lfloor x_0 \rfloor-1=-1$$ Let now $x_0\in \mathbb{Z}$, then : $$\lim_{x\to x_0^+}\lfloor x \rfloor=x_0\;; \; \lim_{x\to x_0^-}\lfloor x \rfloor=x_0-1$$ then \begin{eqnarray} \lim_{x\to x_0^+}f(x)&=&\lim_{x\to x_0^+}(\lfloor x \rfloor +\lfloor -x\rfloor)\\ &=&\lim_{x\to x_0^+}\lfloor x \rfloor +\lim_{x\to x_0^+}\lfloor -x\rfloor\\ &=&\lim_{x\to x_0^+}\lfloor x \rfloor +\lim_{\begin{array}{}x\to x_0\\ x>x_0 \end{array}}\lfloor -x\rfloor\\ &=&\lim_{x\to x_0^+}\lfloor x \rfloor +\lim_{\begin{array}{}-x\to -x_0\\ -x<-x_0 \end{array}}\lfloor -x\rfloor\\ &=&\lfloor x_0 \rfloor+\lim_{-x\to -x_0^-}\lfloor -x \rfloor\\ &=&\lfloor x_0 \rfloor+\lfloor -x_0 \rfloor-1\\ &=&\lfloor x_0 \rfloor-\lfloor x_0 \rfloor-1\\ &=& -1\\ \end{eqnarray} and \begin{eqnarray} \lim_{x\to x_0^-}f(x)&=&\lim_{x\to x_0^-}(\lfloor x \rfloor +\lfloor -x\rfloor)\\ &=&\lim_{x\to x_0^-}\lfloor x \rfloor +\lim_{x\to x_0^-}\lfloor -x\rfloor\\ &=&\lim_{x\to x_0^-}\lfloor x \rfloor +\lim_{\begin{array}{}x\to x_0\\ x<x_0 \end{array}}\lfloor -x\rfloor\\ &=&\lim_{x\to x_0^-}\lfloor x \rfloor +\lim_{\begin{array}{}-x\to -x_0\\ -x>-x_0 \end{array}}\lfloor -x\rfloor\\ &=&\lfloor x_0 \rfloor-1+\lim_{-x\to -x_0^+}\lfloor -x \rfloor\\ &=&\lfloor x_0 \rfloor-1+\lfloor -x_0 \rfloor\\ &=&\lfloor x_0 \rfloor-1-\lfloor x_0 \rfloor\\ &=& -1\\ \end{eqnarray} so $$\lim_{x\to x_0}f(x)=-1 ; \forall x_0\in\mathbb{R}$$ But in $\mathbb{Z}$ $$f(x)=\lfloor x \rfloor+\lfloor -x \rfloor=\lfloor x \rfloor-\lfloor x \rfloor=0$$ so the set $\mathbb{Z}$ is a set of removable discontinuity points of $f$.