\(lim\dfrac{3n^4-4.n^3}{2.n^4-2.n^2}\) là
\(\dfrac{3}{2}\).\(\dfrac{1}{2}\).\(+\infty\).\(-\infty\).Hướng dẫn giải:\(lim\dfrac{3n^4-4.n^3}{2.n^4-2.n^2}=lim\dfrac{3.\dfrac{n^4}{n^4}-4\dfrac{n^3}{n^4}}{2.\dfrac{n^4}{n^4}-2\dfrac{n^2}{n^4}}=\dfrac{3}{2}\).