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 ?
2026-04-06 14:40:02.1775486402
Evaluate $\,\lim\limits_{x \to 0^{-}} \frac{\sin(\lfloor x\rfloor)}{x}$
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Just to the left of $0$, $\sin(\lfloor x \rfloor)$ is about $-.84$. So the left hand limit is $+\infty$.