Does $\lim_{x \to 0} \frac{\sin (\left \lfloor x \right \rfloor)}{\left \lfloor x \right \rfloor}$ exist? 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.
 A: 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.
A: 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. 
