Đáp án:
`lim(\root{3}{n^3-3n^2+1}-sqrt(n^2+4n))=-3`
Giải thích các bước giải:
`lim(\root{3}{n^3-3n^2+1}-sqrt(n^2+4n))`
`=lim[(\root{3}{n^3-3n^2+1}-n)+(n-sqrt(n^2+4n))]`
`=lim[(-3n^2+1)/(\root{3}{(n^3-3n^2+1)^2}+n\root{3}{n^3-3n^2+1}+n^2)+(4n)/(n+sqrt(n^2+4n))]`
`=lim[(-3+1/n^2)/(\root{3}{(1-3/n+1/n^2)^2}+\root{3}{1-3/n+1/n^2}+1)-(4)/(1+sqrt(1+4/n))]`
`=(-3+0)/(\root{3}{(1-0+0)^2}+\root{3}{1-0+0}+1)-(4)/(1+sqrt(1+0))`
`=-1-2=-3`
Vậy `lim(\root{3}{n^3-3n^2+1}-sqrt(n^2+4n))=-3`
