For polynomial $g(x)$ satisfying $(g(a))^2+(g'(a))^2=0$, evaluate $\lim_{x\to a}\frac{g(x)}{g'(x)}\left\lfloor\frac{g'(x)}{g(x)}\right\rfloor$ 
If $g(x)$ is a polynomial function and 
$$(g(\alpha))^2+(g'(\alpha))^2=0$$ then evaluate
$$\displaystyle \lim_{x\rightarrow \alpha}\frac{g(x)}{g'(x)}\bigg\lfloor \frac{g'(x)}{g(x)}\bigg\rfloor $$

Try: from 
$$(g(\alpha))^2+(g'(\alpha))^2=0\quad\implies\quad g(\alpha)=g'(\alpha) = 0 \tag{1}$$
means polynomial $g(x)=0$ has a repeated root, $x=\alpha$.
Using 
$$\frac{g'(x)}{g(x)}-1<\bigg\lfloor\frac{g'(x)}{g(x)}\bigg\rfloor \leq \frac{g'(x)}{g(x)} \tag{2}$$
So
$$\lim_{x\rightarrow \alpha}\bigg(\frac{g'(x)}{g(x)}-1\bigg)\frac{g(x)}{g'(x)}<\lim_{x\rightarrow \alpha}\bigg\lfloor\frac{g'(x)}{g(x)}\bigg\rfloor \frac{g(x)}{g'(x)}\leq \lim_{x\rightarrow \alpha}\frac{g'(x)}{g(x)} \frac{g(x)}{g'(x)} \tag{3}$$
with Squeeze Theorem, the limit must be equal to $1$.
But I have a doubt for left side how can I prove 
$$\displaystyle \lim_{x\rightarrow \alpha}\frac{g(x)}{g'(x)} = 0 \tag{4}$$
Could some help me to explain it? Thanks.

$lim_{x\rightarrow a} \frac{g(x)}{g'(x)}=0$
You can apply  L'Hospital's Rule, and get: $ lim_{x\rightarrow a} \frac{g(x)}{g'(x)}=lim_{x\rightarrow a} \frac{g'(x)}{g''(x)}$
if $g''(a)\neq 0$ you are done, otherwise you can keep going until you get $ lim_{x\rightarrow a} \frac{g^{(n)}(x)}{g^{(n+1)}(x)}=lim_{x\rightarrow a} \frac{g^{(n)}}{c}=\frac{0}{c}$ when $c \neq 0$.
All this is true if $g^{(k)}(x)\neq 0$ for every $x \neq a$ around $a$.
But let say for example $\forall x\in \Bbb R: g(x)=0$ then the limit you are asking for is not defined.
**I am new here, so excuse me if I got some mistakes.. trying my best :)
