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

If $$\mathrm{f}(\mathrm{x})=\mathrm{x}, \quad$$ for $$\mathrm{x} \leq 0$$

$$=0,\quad$$ for $$x>0$$, then the function $$f(x)$$ at $$x=0$$ is

$$\lim _\limits{x \rightarrow 1} \frac{a b^x-a^x b}{x^2-1}=$$

If the function

$$\begin{array}{rlrl} f(x) & =3 a x+b, & & \text { for } x<1 \\ & =11, & & \text { for } x=1 \\ & =5 a x-2 b, & \text { for } x>1 \end{array}$$

is continuous at $$x=1$$. Then, the values of $$a$$ and $$b$$ are