这道函数极限的题,这么做为什么错啊??求大神解答,谢谢!!
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切线方程 (x0, f(x0))
y-f(x0) = f'(x0). ( x-x0)
y=f'(x0). ( x-x0) + f(x0)
y=0
f'(x0). ( x-x0) + f(x0) =0
x = x0 -[f(x0)/f'(x0)]
ie
u= x -[f(x)/f'(x)]
lim(x->0) u/x
=lim(x->0) [ x -f(x)/f'(x) ]/x
=lim(x->0) [ xf'(x) -f(x) ]/[x.f'(x)] (0/0 分子分母分别求导)
=lim(x->0) [ xf''(x) +f'(x) -f'(x) ]/[x.f''(x)+f'(x)]
=lim(x->0) f''(x) /[f''(x)+ f'(x)/x]
=f''(0)/[ f''(0)+f''(0) ]
=1/2
y-f(x0) = f'(x0). ( x-x0)
y=f'(x0). ( x-x0) + f(x0)
y=0
f'(x0). ( x-x0) + f(x0) =0
x = x0 -[f(x0)/f'(x0)]
ie
u= x -[f(x)/f'(x)]
lim(x->0) u/x
=lim(x->0) [ x -f(x)/f'(x) ]/x
=lim(x->0) [ xf'(x) -f(x) ]/[x.f'(x)] (0/0 分子分母分别求导)
=lim(x->0) [ xf''(x) +f'(x) -f'(x) ]/[x.f''(x)+f'(x)]
=lim(x->0) f''(x) /[f''(x)+ f'(x)/x]
=f''(0)/[ f''(0)+f''(0) ]
=1/2
更多追问追答
追问
谢谢,你做对了!但我那么做为什么是错的呢
追答
这个 lim(x->0) f(x)/ [xf'(x)] 你算错!
lim(x->0) f(x)/ [xf'(x)]
=lim(x->0) f(x)/[x f'(x)] (0/0 分子分母分别求导)
=lim(x->0) f'(x)/[f'(x) + xf''(x) ]
=lim(x->0)[ f'(x)/x]/[f'(x)/x + f''(x) ]
=f''(0)/[ f''(0) + f''(0)]
=1/2
1-lim(x->0) f(x)/ [xf'(x)]
=1- 1/2
=1/2
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