Question

如图，求极限，请问结果具体怎么算的？为啥等于1

Answer 1

应该是
lim(n->无穷) n/(n+lnn) = lim(n->无穷) n/(n+ln1) =1
lim(n->无穷) n/(n+ln1)
=lim(n->无穷) n/n
=1
lim(n->无穷) n/(n+lnn)
分子分母同时除n
=lim(n->无穷) 1/(1+lnn/n)
=1/(1+0)
=1
//

lim(n->无穷) e(e^n-1)/[ (e^n+n^2) (e-1) ]
=[e/(e-1) ] lim(n->无穷) (e^n-1)/ (e^n+n^2)
分子分母同时除e^n
=[e/(e-1) ] lim(n->无穷) (1-1/e^n)/ (1+ n^2/e^n)
=[e/(e-1) ] (1-0)/ (1+ 0)
=e/(e-1)
lim(n->无穷) e(e^n-1)/[ (e^n+1^2) (e-1) ]

=lim(n->无穷) e(e^n-1)/[ (e^n+1) (e-1) ]
=[e/(e-1) ] .lim(n->无穷) (e^n-1)/ (e^n+1)
分子分母同时除e^n
=[e/(e-1) ] .lim(n->无穷) (1-1/e^n)/ (1+1/e^n)
=[e/(e-1) ] (1-0)/ (1+ 0)
=e/(e-1)