Question

帮忙解决一道高数题设f(x)具有二阶连续导数,f(0)=0,f '(0)=0,f ''(x)>0.在曲线y=f(x)上任

Answer 1

由f(0)=0,f'(0)=0,f''(x)>0,可知(0,0)是f(x)的全局最小值.因此当x不等于0的时候,f(x)>0,f'(x)不等于0.曲线在点(x,f(x))(x不为0)的切线为z-f(x)=f'(x)(w-x).所以于x轴的截距为u=x-f(x)/f'(x).因此有当x->0时,lim u=0-limf(x)/f'(x)=-limf'(x)/f''(x)=0[洛必达法则].所以当x->0时u->0.当x->0时,根据泰勒展开f(x)~=f(0)+f'(0)x+1/2f''(0)x^2=1/2f''(0)x^2=kx^2,其中k为常数,因此lim xf(u)/uf(x)=lim xku^2/ukx^2=lim u/x=lim (1-f(x)/f'(x)x).对于lim(f(x)/f'(x)x)=lim kx^2/(x(f'(0)+f''(0)x))=lim 1/2=1/2.所以lim xf(u)/uf(x)=1-1/2=1/2.