极限,请问这个怎么求
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lim(x->无穷) { x^3. ln[(x+1)/(x-1)] -2x^2}
y=1/x
=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}
通分母
=lim(y->0+) [ ln[(1+y)/(1-y)] -2y ]/y^3
洛必达
=lim(y->0+) [ 1/(1+y) +1/(1-y) -2 ]/(3y^2)
=lim(y->0+) [ (1-y)+(1+y) -2(1+y)(1-y) ] /[3y^2 .(1+y)(1-y) ]
=(1/3)lim(y->0+) [ (1-y)+(1+y) -2(1+y)(1-y) ] /y^2
=(1/3)lim(y->0+) [ 2 -2(1-y^2) ] /y^2
=(1/3)lim(y->0+) 2y^2 /y^2
=2/3
y=1/x
=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}
通分母
=lim(y->0+) [ ln[(1+y)/(1-y)] -2y ]/y^3
洛必达
=lim(y->0+) [ 1/(1+y) +1/(1-y) -2 ]/(3y^2)
=lim(y->0+) [ (1-y)+(1+y) -2(1+y)(1-y) ] /[3y^2 .(1+y)(1-y) ]
=(1/3)lim(y->0+) [ (1-y)+(1+y) -2(1+y)(1-y) ] /y^2
=(1/3)lim(y->0+) [ 2 -2(1-y^2) ] /y^2
=(1/3)lim(y->0+) 2y^2 /y^2
=2/3
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