高数极限问题设f'(0)＝0,f'(x)在原点的某邻域内连续,且f'(0)≠0,证明lim(x趋近0+)x∧(f(x))＝1

Question

高数极限问题
设f'(0)＝0,f'(x)在原点的某邻域内连续,且f'(0)≠0,证明lim(x趋近0+)x∧(f(x))＝1

Answer 1

题目有错,前面的“设f'(0)＝0”应改为“设f(0)＝0”,否则与后面的“且f'(0)≠0”矛盾.
证：设y=x^(f(x)), 则lny=f(x)lnx
　　∵lim[x-->0+]f(x)lnx
　 　=lim[x-->0+]lnx/(1/f(x))
　 　=-lim[x-->0+][f(x)]^2/[xf'(x)]
　 　=-lim[x-->0+][2f(x)f'(x)]/[f'(x)+xf''(x)]
　 　=0
　　∴lim(x-->0+)x∧(f(x))＝1