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

$$ \begin{array}{r} \lim _{x \rightarrow 0} \frac{2 \tan x+\cos x-1+x}{\sqrt{4 \sin ^2 x+2 \tan x+1}}= \\ -\sqrt{3 \tan ^2 x+\sin x+1} \end{array} $$

If a function $f$ is defined by $f(x)=\frac{\cot ^3 x-\tan x}{\cos (x+\pi / 4)},(x \neq \pi / 4)$, then $\lim _{x \rightarrow \pi / 4} f(x)=$