解:用二重积分的中值定理即可,定理是说:∫∫f(x,)dxdy=f(x0,y0)*S,(x0,y0)为D内某一点,S为积分区域D的面积。本题:∫∫e^(x^2-y^2)cos(x+y)dxdy=[e^(x0^2-y0^2)cos(x0+y0)]*(πr^2),当r趋于0时,点(x0,y0)趋于(0,0),因此所求极限=e^(0^2-0^2)cos(0+0)=1