# Limit $\lim_{x \to 0} \frac{\sin[x]}{[x]}$

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?

• What are RHL, LHL, and $[\cdot]$? – Alex Silva Jan 17 '17 at 9:28
• Right hand limit and left hand limit – Harsh Sharma Jan 17 '17 at 9:29
• I don't know how to prove it, but it seems to not exist. As seen here. – S.C.B. Jan 17 '17 at 9:29
• Quite right, the RHL does not exist, $0/0$ is not defined. But the LHL does. What can you conclude ? – Yves Daoust Jan 17 '17 at 9:29

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