Question

请问一下这个极限怎么求？

Answer 1

这个？

Answer 2

y->0+
ln(1+y)= y-(1/2)y^2 +(1/3)y^3 +o(y^3)
ln(1-y)= -y-(1/2)y^2 -(1/3)y^3 +o(y^3)
ln[(1+y)/(1-y)] - 2y
=ln(1+y)-ln(1-y) -2y
=(2/3)y^3 +o(y^3)
//
y=1/x
lim(x->+∞) { x^3. ln[(x+1)/(x-1)] - 2x^2 }
=lim(y->0+) { (1/y^3). ln[(1/y+1)/(1/y-1)] - 2/y^2 }
=lim(y->0+) { (1/y^3). ln[(1+y)/(1-y)] - 2/y^2 }
=lim(y->0+) { ln[(1+y)/(1-y)] - 2y }/y^3
=lim(y->0+) (2/3)y^3/y^3
=2/3

Answer 3

不会啊，。，。，。，。，。，。，。，。，。，。，。，。，。，。，。，。，，。，。