# 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

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

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 $$-2$$

Because

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

$$Lim=⌊(1/((-1/2)-ϵ))*⌊(-1/((-1/2)-ϵ))⌋⌋$$

$$=⌊(2/(-1-2ϵ)*⌊(-2/(-1-2ϵ)⌋⌋$$

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

therefore $$(-1-2ϵ)<-1$$

and we have

$$Lim=⌊(-2+)*⌊(2−)⌋⌋$$

$$=⌊(-2+)*(1)⌋$$

$$=⌊(-2+)⌋$$

$$=⌊-1.99⌋$$

$$=-2$$

• Please use MathJax for typing math. – Saad Jun 16 '18 at 11:50
• 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 $-(1/2)-\epsilon$. – Barry Cipra Jun 16 '18 at 12:23