TITLE: Limit $\lim_{x \to 0} \frac{\sin[x]}{[x]}$
QUESTION [5 upvotes]: I am trying to find the limit  of  $$\lim_{x \to 0} \frac{\sin[x]}{[x]}$$ where [.] represents the greatest integer function.
I tried to take up an infinitesimally small number $h$ and took up the  Right Hand Limit and Left Hand limit
$$\lim_{x \to 0^+}\frac{\sin[x]}{[x]}$$
$$\Rightarrow \lim_{h \to 0} \frac{\sin[h]}{[h]}$$
I am stuck over here, though I know that $$\lim_{x \to 0}\frac{\sin x}{x}=1$$ But here I see that since $h$ is a very small positive number$[h]$ itself becomes zero and we get
$$\Rightarrow  \frac{\sin 0}{0}.$$ Does this shows that the RHL doesn't exist or am I at fault somewhere?

REPLY [5 votes]: For $-1 \le x<0$, you have $[x]=-1$, so for $x$ in that interval:
$$\frac{\sin[x]}{[x]}=\frac{\sin(-1)}{-1}=\sin 1$$
Which makes the left-handed limit $\sin 1$.
For $0 \le x<1$, you have $[x]=0$ so $\tfrac{\sin x}{x}$ is not defined in neighborhoods on the right of $x=0$.

REPLY [5 votes]: The domain of the function excludes $[0,1)$, so that
$$\lim_{x\to0}f(x)=\lim_{x\to0^-}f(x)=-\sin(-1).$$