Find all values of parameter $a$ for the limit to exist. Let $\left \lfloor {x} \right \rfloor$  to be floor value of $x$.
$$
\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\arcsin \left\lfloor {(a + 2)x} \right\rfloor }}{{x + a}} - \cos \left\lfloor {\left| {ax} \right|} \right\rfloor } \right).
$$
I have problem with this one. Help with direction how to solve it. It seems for me as any $a\neq 0$ is okay. As much as I understand cos we shoulв overlook as there is no uncertainty.
 A: The key observation is that $ \lfloor (a+2)x\rfloor $ and $\lfloor |ax| \rfloor$ both end up being constant for all sufficiently small $x$ with fixed sign, but the first ends at $-1$ or $0$ depending on the sign of $(a+2)x$.  With that in mind, break down into cases:

*

*$a+2>0$ or $a+2 <0$. Then in the first instance,
$$\lfloor(a+2)x\rfloor\to \left\{ \begin{array}{2}0, & \text{when } x \searrow 0 \\
-1, &\text{when } x \nearrow 0 \end{array}\right .$$
and in the second instance the limits are reversed, but continue to differ.
So no limit can exist

*$a+2=0.$  Then $\lfloor(a+2)x\rfloor=0$ for all $x$ and $\lfloor|ax|\rfloor = 0$ for sufficiently small $x$.  The limit then exists and is $\cos(0) =1 $.

Note that $a=0$ is covered in case 1, but it has an additional problem that when $\lfloor (a+2)x \rfloor \to -1$ (one of $x$ and $a+2$ is negative), the left term becomes infinite.
If you restrict $x$ to positive values only then a different conclusion is reached in case 1.   If $(a+2)>0$ the left term eventually becomes zero and you have the same limit as case 2.  If $(a+2)<0$ then the numerator of left term has limit $\arcsin(-1) = -\pi/2$.  As long as $a\neq 0$ the left term then has limit $-\pi/(2a)$.  The right term always has limit $1$.
