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.

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


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

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.

  • $\begingroup$ write $x_0 = n + \epsilon$ where $n \in \mathbb{Z}, \epsilon \in [0,1)$, and consider the case $\epsilon > 0$ and $\epsilon = 0$ $\endgroup$ – reuns Jul 30 '16 at 14:38
  • 2
    $\begingroup$ 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. $\endgroup$ – Elliot G Jul 30 '16 at 14:48
  • $\begingroup$ @ElliotG ah thank you! So the limit doesn't exist since the one-sided limits are not equal to each other? $\endgroup$ – RonaldB Jul 30 '16 at 14:52
  • $\begingroup$ 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 $\endgroup$ – Elliot G Jul 30 '16 at 14:57
  • 1
    $\begingroup$ 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. $\endgroup$ – 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 $f$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.