$$ \text { If } f(x)=\left\{\begin{array}{cc} x & , \quad 0 \leq x \leq 1 \\ 2 x-1 & , \quad x>1 \end{array}\right. \text { then } $$

$$ \text { If } f(x)=\left\{\begin{array}{cc} \frac{1-\sin x}{(\pi-2 x)^2} & , \quad \text { if } x \neq \frac{\pi}{2} \\ \lambda, & \text { if } x=\frac{\pi}{2} \end{array}\right. $$

Then $$f(x)$$ will be continues function at $$x=\frac{\pi}{2}$$, then $$\lambda=$$

$$\lim _\limits{x \rightarrow 0} \frac{\sqrt{a+x}-\sqrt{a}}{x \sqrt{a(a+x)}}$$ equals to

$$ \lim _\limits{x \rightarrow 0}\left(\frac{\sin a x}{\sin b x}\right)^k \text { equals } $$