The function is defined $\mathbb{R}-[ 0,1)\to \mathbb{R}$.

As $x$ approaches $0$ from the left, $\left \lfloor x \right \rfloor=-1$ hence the left hand limit is $\sin \left ( 1 \right )$.

Quite clearly , the right hand limit does not exist.

Now does the limit exist?

On one hand , since LHL is not equal to RHL , it should not exist.

On the other , the definition of limit says

for all $\varepsilon > 0$ , there exists a $\delta > 0 $ such that for all $x $ in $D $ that satisfy $ 0 < | x - c | < \delta $, the inequality $|f(x) - L| < \varepsilon$ holds.

Now since we only consider all $x$ in the domain , I don't see how $x$ not being able to approach from the right creates a problem. I think the definition is still verified if the limit is $\sin \left ( 1 \right )$.

This is what we were told in class (no explanation was given , the definition thing is my idea) but I'm not very sure . Wolfram alpha says the limit does not exist. This question - Find $\lim_{x\to 0}\frac{\lfloor \sin x\rfloor}{\lfloor x\rfloor}$ implies the same.

So please help me. Thank you.

  • $\begingroup$ @Lubin, The OP has specified the domain to be $\mathbb{R}-[0, 1)$, so the function is not defined at $x = 1/3$. $\endgroup$ – mathmandan Feb 21 '15 at 15:49
  • $\begingroup$ (1) "Limit" in the sense used in topology ... in that case, only the domain of the function is used, and your limit is $1$ as you suspect. (2) "Limit" in the sense used in baby calculus books. You will have to check your book carefully, because it may be undefined, or it may be $1$ as before. $\endgroup$ – GEdgar Feb 21 '15 at 16:34
  • $\begingroup$ A similar question would be if the limit of $f(x) = \sqrt{x}$ exists at 0 $\endgroup$ – Alex Feb 21 '15 at 16:57
  • $\begingroup$ @Daniel Fischer , why did you delete your answer? It was good. $\endgroup$ – A Googler Feb 21 '15 at 17:19
  • $\begingroup$ @DanielFischer not sure whether space is necessary. $\endgroup$ – A Googler Feb 21 '15 at 17:20

You can view this two ways. If you are assuming the function ${\sin \lfloor x \rfloor \over \lfloor x \rfloor}$ as a well-defined function on ${\mathbb R} - [0,1)$ then the limit is just the left-hand limit; since the function is defined only to the left of $x = 0$, you only need to take the limit from the left for the limit to exist.

If on the other hand you are just asking if the statement "$\lim_{x \rightarrow 0} {\sin \lfloor x \rfloor \over \lfloor x \rfloor}$ exists" is a true statement, then you can argue that since the function ${\sin \lfloor x \rfloor \over \lfloor x \rfloor}$ is not well-defined on $[0,1)$ the statement has no meaning. But this is not the usual interpretation of such limits. If the function is only defined on one side, usually it is considered to be implicit that you just restrict the domain and consider the limit to be the limit from that side.


Notice that the given function is constant equal $\sin(1)$ on a neighborhood of $0$ (i.e. where $0$ is a limit point of this neighborhood) so the limit of the funtion exists and it's equal to the mentioned constant.

  • $\begingroup$ Is this true even if domain is not explicitly mentioned? Consider this probem: math.stackexchange.com/questions/364552/… Shouldn't that limit also exist as we don't consider the points at which function is undefined? But the answer with 3 upvotes says it doesn't and the person who has written the accepted answer also agrees that it doesn't exist (int he comments to that answer). $\endgroup$ – A Googler Feb 21 '15 at 15:57
  • $\begingroup$ Even in the link that you gave the limit exists: the reason is that the neighborhood that we consider isn't necessarily an interval so IMHO the answer with 3 upvotes is wrong. $\endgroup$ – user63181 Feb 21 '15 at 16:05

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