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 !

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.
