设曲线f(x)=x^2n在点(1,1)处的切线与x轴的交点为(An,0),求limx->无穷f(An)

Question

设曲线f(x)=x^2n在点(1,1)处的切线与x轴的交点为(An,0),求limx->无穷f(An)

Answer 1

还是去年的,过了快一个月了
f'(x)=2n*x^(2n-1)
斜率kn=f'(1)=2n
切线ln: y-1=kn(x-1)
令y=0得,xn=1-1/kn=1-1/(2n)
即An=1-1/(2n)
f(An)=[1-1/(2n)]^(2n)
lim(n->∞)f(An)
=lim(n->∞)[1-1/(2n)]^(2n)
=lim(n->∞)1/[1+1/(-2n)]^(-2n)
=1/e