设x→0,limf(x)/x=0,f''(0)=4,证明:x→0,limf(x)/x^2=2

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设x→0,limf(x)/x=0,f''''(0)=4,证明:x→0,limf(x)/x^2=2设x→0,limf(x)/x=0,f''''(0)=4,证明:x→0,limf(x)/x^2=2设x→0,lim

设x→0,limf(x)/x=0,f''(0)=4,证明:x→0,limf(x)/x^2=2
设x→0,limf(x)/x=0,f''(0)=4,证明:x→0,limf(x)/x^2=2

设x→0,limf(x)/x=0,f''(0)=4,证明:x→0,limf(x)/x^2=2
x→0
因为f''(0)=4,故,f'(x)在x=0处连续

lim f(x)/x极限存在,故该极限必为0/0型,利用L'Hospital法则
=lim f'(x)
=f'(0)
=0
lim f(x)/x^2
该极限为0/0型,利用L'Hospital法则,
=lim f'(x)/2x
该极限为0/0型,利用L'Hospital法则,
=lim f''(x)/2
=f''(0)/2
=4/2
=2
有不懂欢迎追问

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