text: # 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