a) \(\lim_{x\to\infty}\frac{-3n^3+2n^2+1}{2n^3-1}\)
\(=\lim_{x\to\infty}\frac{n^3\left(-3+\frac{2}{n}+\frac{1}{n^3}\right)}{n^3\left(2-\frac{1}{n^2}\right)}\)
\(\) \(=\lim_{x\to\infty}\frac{-3+\frac{2}{n^2}_{}+\frac{1}{n^3}}{n^3\left(2-\frac{1}{n^2}\right)}=\frac{-3}{2}\)
b)\(\lim_{x\to\infty}\frac{2n+1}{n^2+n+3}=\lim_{x\to\infty}\frac{n\left(2+\frac{1}{n}\right)}{n^2\left(1+\frac{1}{n}+\frac{3}{n^2}\right)}\)
\(=\lim_{x\to\infty}\frac{n}{n^2}\frac{2+\frac{1}{n}}{1+\frac{1}{n}+\frac{3}{n^2}}\)
\(=\lim_{x\to\infty}\frac{1}{n^{}}.\lim_{x\to\infty}\frac{2+\frac{1}{n}}{1+\frac{1}{n}+\frac{3}{n^2}}\)
\(=0.\lim_{x\to\infty}\frac{2+\frac{1}{n}}{1+\frac{1}{n}+\frac{3}{n^2}}=0\)
c)\(\lim_{x\to\infty}\frac{2025n-2024}{n-2023}=\lim_{x\to\infty}\frac{n\left(2025-2024\frac{1}{n}\right)}{n\left(1-\frac{2023.1}{n}\right)}\)
\(=\lim_{x\to\infty}\frac{2025-\frac{2024}{n}}{1-\frac{2023}{n}}\)
\(=\frac{2025-2024.0}{1-2023.0}=\frac{2025}{1}=2025\)
