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?


3 Answers 3


Assume $x=-0.5-\epsilon$ with $\epsilon >0$, then we have that


therefore for $\epsilon$ sufficiently small we have

$$\left\lfloor\frac{1}{x} \left\lfloor \frac{-1}{x} \right\rfloor\right\rfloor=\left\lfloor (-2+4\epsilon) \left\lfloor 2-4\epsilon \right\rfloor\right\rfloor=\left\lfloor (-2+4\epsilon) \right\rfloor=-2$$


Start by looking at the interior floor function. As $h \rightarrow 0^+$, we know that $(2-h)$ is slightly smaller than $2$, so $\lfloor (2-h) \rfloor$ will always evaluate to $1$. Similarly, $(-2+h)$ will be slightly greater than $-2$. After we multiply by $1$ (the evaluation of $\lfloor (2-h) \rfloor$), we can take the floor to get $-2$ for every small positive $h$.


The correct answer is $-2$


when ($x→−0.5^{-}$) therefore $x=−0.5−ϵ$ with $ϵ>0$, then we have that



We know that $ϵ$ is a very small positive number

therefore $(-1-2ϵ)<-1$

and we have





$ =-2$

  • $\begingroup$ Please use MathJax for typing math. $\endgroup$ Jun 16, 2018 at 11:50
  • $\begingroup$ You dropped the minus sign when you inserted $x=-0.5-\epsilon$ into the floor functions as $(1/2)-\epsilon$. You need to insert it as $-(1/2)-\epsilon$. $\endgroup$ Jun 16, 2018 at 12:23

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