$$\mathop {\lim }\limits_{x \to 0}\left(\frac{4!}{x^8}\left(1-\cos \frac{x^2}{3}-\cos \frac{x^2}{4}+\cos \frac{x^2}{3} \cos \frac{x^2}{4}\right)\right)= $$

Let $A=\left(a_{i j}\right)$ be an $n \times n$ matrix defined by $a_{i j}=\left\{\begin{array}{cc}k^i, & \forall i=j \\ 0, & \text { otherwise }\end{array}\right.$. If $m=$ trace of $A$ and $\lim _{k \rightarrow 1} \frac{n-m}{1-k}=171$, then the value of $n$ is

$$\mathop {\lim }\limits_{x \to \infty } {x^3}\left[\sqrt{x^2+\sqrt{x^4+1}}-\sqrt{2 x}\right]= $$

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, $\alpha+\beta=$