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

$\lim _{x \rightarrow 0} \frac{3^{\sin x}-2^{\tan x}}{\sin x}=$

If the function

$$ f(x)=\left\{\begin{array}{cc} \frac{\cos a x-\cos 9 x}{x^2} & \text {, if } x \neq 0 \\ 16 & \text {, if } x=0 \end{array}\right. $$

is continuous at $x=0$, then $a=$

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