(1+2+3++.+n)/(n+3)(n+4)的极限怎么样做呀
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(1+2+3++.+n)/(n+3)(n+4)的极限怎么样做呀
![(1+2+3++.+n)/(n+3)(n+4)的极限怎么样做呀](/uploads/image/z/7621149-21-9.jpg?t=%281%2B2%2B3%2B%2B.%2Bn%29%2F%28n%2B3%29%28n%2B4%29%E7%9A%84%E6%9E%81%E9%99%90%E6%80%8E%E4%B9%88%E6%A0%B7%E5%81%9A%E5%91%80)
(1+2+3++.+n)/(n+3)(n+4)= 分子求和
=n(1+n)/(n+3)(n+4) 分子分母同时除以n^2
lim[n(1+n)/(n+3)(n+4)](n->+无穷)=lim[(1/n+1)/(1+7/n+12/n^2)](n->+无穷)=1
=n(1+n)/(n+3)(n+4) 分子分母同时除以n^2
lim[n(1+n)/(n+3)(n+4)](n->+无穷)=lim[(1/n+1)/(1+7/n+12/n^2)](n->+无穷)=1
(1+2+3++.+n)/(n+3)(n+4)的极限怎么样做呀
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