设函数f(x)在x=0的某邻域具有一阶连续导数,且f(0)f′(0)≠0,当h→0时,若af(h)+bf(2h)-f(0)=0(h),试求a,b的值.
问题描述:
设函数f(x)在x=0的某邻域具有一阶连续导数,且f(0)f′(0)≠0,当h→0时,若af(h)+bf(2h)-f(0)=0(h),试求a,b的值.
答
由题设条件知:limh→0[af(h)+bf(2h)−f(0)]h=limh→0(a+b−1)f(0)h=0,∴(a+b-1)f(0)=0,由于:f(0)f′(0)≠0,故必有:a+b-1=0.…①又由洛必达法则知:limh→0af(h)+bf(2h)−f(0)h=limh→0af′(h)+2bf...