\(C1:\lim\limits\dfrac{3n^2-2n^4-3}{-n^3+4n+5}\\=\lim\limits\dfrac{n^4\left(\dfrac{3}{n^2}-2-\dfrac{3}{n^4}\right)}{n^3\left(-1+\dfrac{4}{n^2}+\dfrac{5}{n^3}\right)}\\=\lim\limits n.\dfrac{\dfrac{3}{n^2}-2-\dfrac{3}{n^4}}{-1+\dfrac{4}{n^2}+\dfrac{5}{n^3}}\\=\lim\limits n.\lim\limits \dfrac{\dfrac{3}{n^2}-2-\dfrac{3}{n^4}}{-1+\dfrac{4}{n^2}+\dfrac{5}{n^3}}\\\text{Ta có:}\lim\limits n=+\infty;\lim\limits\dfrac{\dfrac{3}{n^2}-2-\dfrac{3}{n^4}}{-1+\dfrac{4}{n^2}+\dfrac{5}{n^3}}=\dfrac{0-2-0}{-1+0+0}=2.\\\Rightarrow \lim\limits n.\lim\limits\dfrac{\dfrac{3}{n^2}-2-\dfrac{3}{n^4}}{-1+\dfrac{4}{n^2}+\dfrac{5}{n^3}}=+\infty.2=+\infty\\\Rightarrow \lim\limits\dfrac{3n^2-2n^4-3}{-n^3+3n+5}=+\infty\\C2:\lim\limits\dfrac{3n^2-2n^4-3}{-n^3+4n+5}\\=\lim\limits\dfrac{\dfrac{3n^2}{n^4}-\dfrac{2n^4}{n^4}-\dfrac{3}{n^4}}{-\dfrac{n^3}{n^4}+\dfrac{4n}{n^4}+\dfrac{5}{n^4}}\\=\lim\limits\dfrac{\dfrac{3}{n^2}-2-\dfrac{3}{n^4}}{-\dfrac{1}{n}+\dfrac{4}{n^3}+\dfrac{5}{n^4}}\\=\lim\limits\dfrac{0-2-0}{0+0+0}=+\infty.\\\text{C3:Áp dụng phương pháp vô cùng lớn:}\\\lim\limits\dfrac{3n^2-2n^4-3}{-n^3+4n+5}\\=\lim\limits\dfrac{-2n^4}{-n^3}\\=\lim\limits 2n=+\infty.\)
