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

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, $\alpha+\beta=$

If $a f(x)+b f\left(\frac{1}{x}\right)=x+1$, and $\frac{d}{d x}\left(x^2 f(x)\right)=2 x^2+2 x+\frac{1}{3}$, then $a-b$

If $f(x)=\sin \left(\cosh \left(\frac{x^2+1}{x^2+2}\right)\right)$, then $f^{\prime}(1)=$