Let $f(x)=\left\{\begin{array}{ccc}3-x & \text { if } & x<-3 \\ 6 & \text { if } & -3 \leq x \leq 3 . \text { Let } \alpha \text { be the number } \\ 3+x & \text { if } & x>3\end{array}\right.$ of points of discontinuity of $f$ and $\beta$ be the number of points where $f$ is not differentiable. Then, $\alpha+\beta=$

$$ \lim _{x \rightarrow 3^{-}} \frac{x^3-3 x^2-4 x+12}{2 x^3-7 x^2+2 x+3}= $$

$$ \lim _{x \rightarrow 0} \frac{2^{2 x}-2^{x+1}+2-\cos 2 x}{x^2}= $$

If $f(x)=\left\{\begin{array}{l}\frac{x^2-16}{x-4} \text { if } x>4 \\ 2 x \quad \text { if } x \leq 4\end{array}\right.$ then $f^{\prime}\left(4^{-}\right)+f^{\prime}\left(4^{+}\right)=$