If $\lim \limits_{x \rightarrow 4} \frac{2 x^2+(3+2 a) x+3 a}{x^3-2 x^2-23 x+60}=\frac{11}{9}$, then $\lim \limits_{x \rightarrow a} \frac{x^2+9 x+20}{x^2-x-20}=$

If the function $$ f(x)= \begin{cases}\frac{\tan a(x-1)}{x-1}, & \text { if } 04\end{cases} $$ domain, then $6 a+9 b^4=$

$\lim _{\theta \rightarrow \frac{\pi^{-}}{2}} \frac{8 \tan ^4 \theta+4 \tan ^2 \theta+5}{(3-2 \tan \theta)^4}=$

Define $ f: R \rightarrow R $ by $ f(x)=\left\{\begin{array}{cl}\frac{1-\cos 4 x}{x^{2}}, & x < 0 \\ a, & x=0 \\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & x > 0\end{array}\right. $

Then, the value of $ a $ so that $ f $ is continuous at $ x=0 $ is