f(x)=√x / [(1+x)^1/2+1] ,f(0)=01.lim[ f(x),x->0+] = 0 = f(0) 即 f(x)在x=0点右连续2.f '+(0) = lim[ [f(x)-f(0)] / (x-0),x->0+] = lim[ [(1+x)^1/2+1] /√x,x->0+] = ∞即 f(x)在x=0点的右导数不存在...