Basic question about floor function and limit ( $\lim \limits_{x \to 0} \lfloor{x-2}\rfloor \cdot \lfloor{x+3}\rfloor$) 
$\lim \limits_{x \to 0} \lfloor{x-2}\rfloor \cdot \lfloor{x+3}\rfloor$
calculate the limit if it exists if not then prove it does not exist


I tried approaching by squeeze theorem and floor function property and got
$(x-2) \cdot (x+3)-1<\lfloor{x-2}\rfloor \cdot \lfloor{x+3}\rfloor \leq (x-2) \cdot (x+3)$
but then if I calculate the limits as $x$ approaches zero I get
$-7<\lfloor{x-2}\rfloor \cdot \lfloor{x+3}\rfloor \leq-6$
which did not give me an answer according to squeeze theorem so I tried a different approach by side limits
$\lim \limits_{x \to 0^+} \lfloor{x-2}\rfloor \cdot \lfloor{x+3}\rfloor = \lfloor{0-2}\rfloor \cdot \lfloor{0+3}\rfloor = -6$
and $\lim \limits_{x \to 0^-} \lfloor{x-2}\rfloor \cdot \lfloor{x+3}\rfloor = \lfloor{-1-2}\rfloor \cdot \lfloor{-1+3}\rfloor = -6$
so the limit exists and $L=-6$
is this correct? is there a different way?
thank you !
 A: As $x$ approaches $0$ from above (i.e. the right side limit), you have that
$\lfloor x-2\rfloor~$ stays at $~-2~$ and 
$\lfloor x+3\rfloor~$ stays at $~3.~$ 
Therefore, the product stays at $~-6.~$
As $x$ approaches $0$ from below (i.e. the left side limit), you have that
$\lfloor x-2\rfloor~$ stays at $~-3~$ and 
$\lfloor x+3\rfloor~$ stays at $~2.~$ 
Therefore, the product stays at $~-6.~$
So, the limit, as $x$ approaches $0$ from above does in fact equal the limit as $x$ approaches $0$ from below, and this limit is $-6.$

What makes this problem unusual is that you have the limit of the product of two functions, $~\lfloor x-2\rfloor~$ and $~\lfloor x+3\rfloor,~$ where for each function, as $x$ approaches $0$, the left side limit of the function is not equal to the right side limit of the function.
Despite that, when examining the product of the two functions, as $x$ approaches $0$, the left side limit of the product does equal the right side limit of the product.
A: $x+2 < \lfloor x+3\rfloor\leq x+3$ and $ x-3<\lfloor x-2 \rfloor\leq x-2$.
this implies $(x+2).(x-3)<\lfloor x+3 \rfloor.\lfloor x-2 \rfloor\leq (x+3).(x-2)$
so $lim_{x \to 0}(x+2).(x-3)<lim_{x \to 0}\lfloor x+3 \rfloor.\lfloor x-2 \rfloor\leq lim_{x \to 0} (x+3).(x-2)$
$\Rightarrow $$lim_{x \to 0}\lfloor x+3 \rfloor.\lfloor x-2 \rfloor=-6.$
warning : the first implication is not necessary true because if the both left sides are negative   and the other are positive maybe in one case we must reverse the product of sides  ,but any way in all cases it will get the same limit $-6$.so try to  rewrite the complete proof by yourself .
