Evaluate $\,\lim\limits_{x \to 0^{-}} \frac{\sin(\lfloor x\rfloor)}{x}$

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How can I evaluate this limit. $$\lim_{x \to 0^{-}} \frac{\sin(\lfloor x\rfloor)}{x}$$ This is what I did:
$x\in \big(\frac{-1}{2},0\big)\;$ ; $\;\;\lfloor x\rfloor =-1 $ $$\lim_{x \to 0^{-}}\sin(\lfloor x\rfloor)=\sin(-1)<0 \;\;\text{ and }\;\lim_{x \to 0^{-}} x =0^{-}$$ then
$$\lim_{x \to 0^{-}} \frac{\sin(\lfloor x\rfloor)}{x}=+\infty$$ What you think? Any other method ?

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Just to the left of $0$, $\sin(\lfloor x \rfloor)$ is about $-.84$. So the left hand limit is $+\infty$.