大师作文网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),
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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
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
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