Question

请问这个极限怎么求

设f(x)具有二阶连续导数，f(0)=0,f'(0)=0,f''(0)>0,在曲线y=f(x)上任意一点（x,f(x))（x≠0）处作此曲线的切线，此切线在X轴上的截距记为u,求limxf(u)/uf(x),x->0

Answer 1

由题意可以知道 u = x - f(x)/f'(x)
f(0)=0,f'(0)=0,f''(0)>0
设f(x) = x²(a+bx+cx²+......)
那么f(u) = u²(a+bu+cu²+......)
所以xf(u)/uf(x) = u(a+bx+cx²+....) / (x(a+bu+cu²+......))
x->0，那么u->0
所以limxf(u)/uf(x) = lim(u/x) =lim ( 1- f(x) / xf'(x) )
f(x) / xf'(x) 是0/0型的极限，所以
洛必达法则得lim f(x) / xf'(x) =lim f'(x) / [f'(x) + xf''(x)] = lim f''(x) / [f''(x) + f''(x) + xf'''(x) ] = f''(0) / (2f''(0) + 0) = 1/2
所以limxf(u)/uf(x) = lim(u/x) =lim ( 1- f(x) / xf'(x) ) = 1-1/2 = 1/2

Answer 2

由题可知，f(x)=ax²+o(x²)
u=x-f(x)/f'(x)
lim u/x=lim [1-f(x) / xf'(x)]
而lim f(x) / xf'(x) =lim f'(x) / [f'(x) + xf''(x)] = lim f''(x) / [f''(x) + f''(x) + xf'''(x) ] = f''(0) / (2f''(0) + 0) = 1/2，所以lim u/x=1/2
且lim f(u)/f(x)=lim [au²+o(u²)]/[ax²+o(x²)]=lim u²/x²=1/4
所以原式=[lim f(u)/f(x)]/(lim u/x)=1/2

Answer 3

2. (x趋向于0) =limx^3/(x-sinx) =lim3x^2/(1-cosx) =lim6x/sinx[ln(lnx)]