Evaluate $ \lim_{x \to -0.5^{-}} \left\lfloor\frac{1}{x} \left\lfloor \frac{-1}{x} \right\rfloor\right\rfloor$ Evaluate $$L=\lim_{x \to -0.5^{-}} \left\lfloor\frac{1}{x} \left\lfloor \frac{-1}{x} \right\rfloor\right\rfloor $$
My try:
Let $t=\frac{1}{x}$ Now when $ t \to -0.5^{-}$ we have $t \to -2^{+}$ we get
$$L=\lim_{t \to -2^{+}}  \left\lfloor t \left\lfloor -t \right\rfloor \right\rfloor =\lim_{h \to 0}\left\lfloor (-2+h) \left\lfloor (2-h) \right\rfloor \right\rfloor$$
How can we proceed now since we cannot take limit inside greatest integer function?

Assume $x=-0.5-\epsilon$ with $\epsilon >0$, then we have that
$$\frac1x=\frac1{-0.5-\epsilon}=-\frac{2}{1+2\epsilon}=-2(1-2\epsilon)=-2+4\epsilon$$
therefore for $\epsilon$ sufficiently small we have
$$\left\lfloor\frac{1}{x} \left\lfloor \frac{-1}{x} \right\rfloor\right\rfloor=\left\lfloor (-2+4\epsilon) \left\lfloor 2-4\epsilon \right\rfloor\right\rfloor=\left\lfloor (-2+4\epsilon) \right\rfloor=-2$$
