$$ \mathop {\lim }\limits_{n \to \infty } \frac{1}{n^3} \sum\limits_{k=1}^n k^2 x= $$

Let $f: R \rightarrow R$ be defined by

$$ f(x)=\left\{\begin{array}{cc} a-\frac{\sin [x-1]}{x-1}, & \text { if } x>1 \\ 1, & \text { if } x=1 \\ b-\left[\frac{\sin [x-1]-[x-1]}{([x-1])^3},\right. & \text { if } x<1 \end{array}\right. $$

where $[t]$ denotes the greatest integer less than or equal to $t$. If $f$ is continuous at $x=1$, then $a+b=$

$$ \mathop {\lim }\limits_{y \to 0} \frac{\sqrt{1+\sqrt{1+y^4}}-\sqrt{2}}{y^4}= $$

If $\mathop {\lim }\limits_{x \to 0} \frac{\cos 2 x-\cos 4 x}{1-\cos 2 x}=k$, then $\lim\limits_{x \rightarrow k} \frac{x^k-27}{x^{k+1}-81}=$