I am unable to evaluate this limit. The floor function is giving me trouble. Any help will be appreciated. And please edit it so that it looks readable.
2026-03-27 15:18:59.1774624739
Find $\lim_{x\to 0}\frac{\lfloor \sin x\rfloor}{\lfloor x\rfloor}$
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For $0<x<1$ we have $\lfloor x\rfloor=0$ so the fraction isn't defined so we can only look for the left limit
$$\lim_{x\to 0^-}\frac{\lfloor \sin x\rfloor}{\lfloor x\rfloor}=\frac{-1}{-1}=1$$