The value of $$\lim _\limits{x \rightarrow 0} \frac{e^{a x}-e^{b x}}{2 x}$$ is equal to

If $$f(x) = \left\{ {\matrix{ {2\sin x} & ; & { - \pi \le x \le {{ - \pi } \over 2}} \cr {a\sin x + b} & ; & { - {\pi \over 2} < x < {\pi \over 2}} \cr {\cos x} & ; & {{\pi \over 2} \le x \le \pi } \cr } } \right.$$ and it is continuous on $$[-\pi, \pi]$$, then

The value of $$\lim _\limits{x \rightarrow \infty}\left(\frac{x^2-2 x+1}{x^2-4 x+2}\right)^{2 x}$$ is

$$ \lim _\limits{x \rightarrow 0} \frac{a^x-b^x}{x} \text { is equal to } $$