Find the right hand limit of the given function

$$\lim_{x\to 0^+}\frac{\sin [x]}{[x]}$$,Where $[.]$ denotes greatest integer function.

My Attempt:

I just expanded the $\sin $ function then divided it by $[x]$ Then taken the limit and found the limit as $1$, But I am not sure about my solution. Please someone help me. Thank you.

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    $\begingroup$ The function is not defined for any positive $x$ less than $1$. $\endgroup$ – André Nicolas Sep 14 '15 at 15:17
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    $\begingroup$ @André Nicolas Sir can you explain more Please. why it is undefined? for $x\in (0,1), [x]=0$ so we will get $0/0$. is it right? $\endgroup$ – aryan Sep 14 '15 at 15:27
  • $\begingroup$ @aryan It's undefined because 0/0 is undefined. BTW, your notation is confusing. Does sin[x] mean the sin of he greatest integer function? $\endgroup$ – Jerry Guern Sep 14 '15 at 15:33
  • $\begingroup$ @JerryGuern Sir yes sin[x] mean the sin of he greatest integer function $\endgroup$ – aryan Sep 14 '15 at 15:34
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    $\begingroup$ I assume you are asking about $\frac{\sin(\lfloor x\rfloor)}{\lfloor x\rfloor}$, as $x$ approaches $0$ from the right. If you set $x$, for example, equal to $1/10$, on top you will have $\sin(0)$, that is, $0$. On the bottom you will also have $0$. So the ratio does not exist. The situation is different if we are approaching $0$ from the left. $\endgroup$ – André Nicolas Sep 14 '15 at 16:10

In order for a right-hand limit $\lim\limits_{x \to a^{+}} f(x)$ to make sense, there must exist a $\delta > 0$ such that the function $f$ is defined in the open interval $(a, a + \delta)$.

Since (with $k$ denoting an integer) $$ f(x) = \frac{\sin [x]}{[x]} = \begin{cases} \dfrac{\sin k}{k} & k \leq x < k + 1,\ k \neq 0, \\ \text{undefined} & 0 \leq x < 1, \end{cases} $$ the right-hand limit of $f$ at $0$ makes no sense.

(The dashed line in the plot is the graph $y = \sin x/x$ for $x \neq 0$.)

The graph of $\sin[x]/[x]$

That said, it's conceivable the question (somewhat perversely) refers to the continuous extension $$ g(x) = \begin{cases} \dfrac{\sin x}{x} & x \neq 0, \\ 1 & x = 0, \end{cases} $$ and that $f(x) = g([x])$. If this interpretation were correct, $$ f(x) = \frac{\sin [x]}{[x]} = \begin{cases} \dfrac{\sin k}{k} & k \leq x < k + 1,\ k \neq 0, \\ 1 & 0 \leq x < 1, \end{cases} $$ and $\lim\limits_{x \to 0^{+}} f(x) = 1$.


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