f(x)在x=0的领域内有二阶导数,又x→0时lim((sinx+xf(x))\x3)=1/2,求f(0),f'(0),f''(0)
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/24 01:30:51
![f(x)在x=0的领域内有二阶导数,又x→0时lim((sinx+xf(x))\x3)=1/2,求f(0),f'(0),f''(0)](/uploads/image/z/5437796-68-6.jpg?t=f%28x%29%E5%9C%A8x%3D0%E7%9A%84%E9%A2%86%E5%9F%9F%E5%86%85%E6%9C%89%E4%BA%8C%E9%98%B6%E5%AF%BC%E6%95%B0%2C%E5%8F%88x%E2%86%920%E6%97%B6lim%28%28sinx%2Bxf%28x%29%29%5Cx3%29%3D1%2F2%2C%E6%B1%82f%280%29%2Cf%27%280%29%2Cf%27%27%280%29)
f(x)在x=0的领域内有二阶导数,又x→0时lim((sinx+xf(x))\x3)=1/2,求f(0),f'(0),f''(0)
f(x)在x=0的领域内有二阶导数,又x→0时lim((sinx+xf(x))\x3)=1/2,求f(0),f'(0),f''(0)
f(x)在x=0的领域内有二阶导数,又x→0时lim((sinx+xf(x))\x3)=1/2,求f(0),f'(0),f''(0)
根据洛笔答法则,
lim((sinx+xf(x))/x3)=lim((cosx+f(x)+x·f'(x))/3x²)
若x→0时这个极限存在,则必有lim cosx+f(x)+x·f'(x)=0
则cos0+f(0)=0
f(0)=-1
再进一步用洛笔答法则得
lim((cosx+f(x)+x·f'(x))/3x²)
=lim((-sinx+2f'(x)+x·f''(x))/6x)
若x→0时这个极限存在,则必有lim -sinx+2f'(x)+x·f''(x)=0
则f'(0)=0.
则
lim((-sinx+2f'(x)+x·f''(x))/6x)
=(1/6) [lim(-sinx /x) +2lim f'(x)/x +lim f''(x)]
=(1/6) [-1 +2lim (f'(x)-f'(0))/(x-0) + f''(0)]
=(1/6) [-1 +2f''(0) + f''(0)]
=(1/6) [-1 +3f''(0)]
即(1/6) [-1 +3f''(0)]=1/2.
则f''(0)=4/3