If $ f(x)=\left\{\begin{array}{ll}\frac{8}{x^{3}}-6 x & \text {, if } 0 < x \leq 1 \\\\ \frac{x-1}{\sqrt{x}-1} & \text {,if } x > 1\end{array}\right. $ is a real valued function, then at $ x=1, f $ is

$$ \lim\limits_{x \rightarrow 1}(1-x) \tan \left(\frac{\pi}{2} x\right)= $$

If $f(9)=9$ and $f^{\prime}(9)=4$, then $\lim\limits_{x \rightarrow 9} \frac{\sqrt{f(x)}-3}{\sqrt{x}-3}=$