掌握一个准则：可导必连续
可导必连续，在分界点 x = 0 处左极限是 limf(x) = lim[1-√(1-x)]/x = lim1/[1+√(1-x)] = 1/2右极限是 limf(x) = lima+bx = a = f(0), 故得 a = 1/2.左导数是 lim[f(x)-f(0)]/x = lim{[1-√(1-x)]/x-1/2}/x= (1/2)lim[2-2√(1-x)-x]/x^2 (0/0)= (1/2)lim[1/√(1-x)-1]/(2x) = (1/4)lim[1-√(1-x)]/[x√(1-x)]= (1/4)lim1/{√(1-x)[1+√(1-x)]} = 1/8右导数是 lim[f(x)-f(0)]/x = limbx/x = b, 故得 b = 1/8.
设g(x)=(1-√1-x)/x，x∈R则lim(x--＞0)g(x)=1/2g'(0)=1/4所以f(0)=1/2 即a=1/2f'(0)=g'(0) 即b=1/4
