Question

已知函数f(x)连续,且lim(x->0)[f(x)/x]=2,则曲线y=f(x)应对x=0处切线方程为?

Answer 1

lim(x->0)[f(x)/x]≠∞ →f(2)=0

lim(x->0)[f(x)/x]=lim(x->0)[ ( f(x) -f(0) ) / (x-0) ] =f'(2)=2

说明在x=0处切线斜率为2

则切线方程为y=2x