[tex]\displaystyle\\\lim_{n\to\infty}\left(\dfrac{4n-n^2+3n^4}{5-n^3-2n^4}\right)^3=\lim_{n\to\infty}\left(\dfrac{n^4\left(\dfrac{4}{n^3}-\dfrac{1}{n^2}+3\right)}{n^4\left(\dfrac{5}{n^4}-\dfrac{1}{n}-2\right)}\right)^3=\lim_{n\to\infty}\left(\dfrac{\dfrac{4}{n^3}-\dfrac{1}{n^2}+3}{\dfrac{5}{n^4}-\dfrac{1}{n}-2}\right)^3=\\=\dfrac{0-0+3}{0-0-2}=\dfrac{3}{-2}=-\dfrac{3}{2}[/tex]
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[tex]\displaystyle\\\lim_{n\to\infty}\left(\dfrac{4n-n^2+3n^4}{5-n^3-2n^4}\right)^3=\lim_{n\to\infty}\left(\dfrac{n^4\left(\dfrac{4}{n^3}-\dfrac{1}{n^2}+3\right)}{n^4\left(\dfrac{5}{n^4}-\dfrac{1}{n}-2\right)}\right)^3=\lim_{n\to\infty}\left(\dfrac{\dfrac{4}{n^3}-\dfrac{1}{n^2}+3}{\dfrac{5}{n^4}-\dfrac{1}{n}-2}\right)^3=\\=\dfrac{0-0+3}{0-0-2}=\dfrac{3}{-2}=-\dfrac{3}{2}[/tex]