Lời giải:
$limd_{n}=lim\frac{(2-3n)^3.(n+1)^2}{1-4n^5}$
$=lim\frac{(8-36n+54n^2-27n^3)(n^2+2n+1)}{1-4n^5}$
$=lim\frac{8n^2+16n+8-36n^3-72n^2-36n+54n^4+108n^3+54n^2-27n^5-54n^4-27n^3}{1-4n^5}$
$=lim\frac{-27n^5+45n^3-10n^2-20n+8}{1-4n^5}$
$=lim\frac{-27n^5.(1-\frac{45}{27n^2}+\frac{10}{27n^3}+\frac{20}{27n^4}-\frac{8}{27n^5})}{-4n^5.(-\frac{1}{4n^5}+1)}$
$=\frac{27}{4}$