lim ((n – 1) ^ 2 * (3n + 1))/(n ^ 3 + 3n – 1)

2 bình luận về “lim ((n – 1) ^ 2 * (3n + 1))/(n ^ 3 + 3n – 1)”

1. lim {(n – 1)^2. (3n + 1)}/{n^3 + 3n – 1}$\\$

   = lim {(n^2 – 2n + 1). (3n + 1)}/{n^3 + 3n – 1}$\\$

   = lim {3n^3 – 5n^2 + n + 1}/{n^3 + 3n – 1}$\\$

   = lim {n^3. (3 – 5/{n} + 1/{n^2} + 1/{n^3})}/{n^3. (1 + 3/{n^2} – 1/{n^3})}$\\$

   = lim {3 – 5/{n} + 1/{n^2} + 1/{n^3}}/{1 + 3/{n^2} – 1/{n^3}}$\\$

   = 3 $\\$

    

    

   Trả lời
2. Giải đáp:

   $\text{lim}\dfrac{(n-1)^2.(3n+1)}{n^3+3n-1}=3$

   Lời giải và giải thích chi tiết:

   $\text{lim}\dfrac{(n-1)^2.(3n+1)}{n^3+3n-1}\\=\text{lim}\dfrac{(n^2-2n+1)(3n+1)}{n^3+3n-1}\\=\text{lim}\dfrac{3n^3+n^2-6n^2-2n+3n+1}{n^3+3n-1}\\=\text{lim}\dfrac{3n^3-5n^2+n+1}{n^3+3n-1}\\=\text{lim}\dfrac{n^3\Bigg{(}3-\dfrac{5}{n}+\dfrac{1}{n^2}+\dfrac{1}{n^3}\Bigg{)}}{n^3\Bigg{(}1+\dfrac{3}{n^2}-\dfrac{1}{n^3}\Bigg{)}}\\=\text{lim}\dfrac{\Bigg{(}3-\dfrac{5}{n}+\dfrac{1}{n^2}+\dfrac{1}{n^3}\Bigg{)}}{\Bigg{(}1+\dfrac{3}{n^2}-\dfrac{1}{n^3}\Bigg{)}}\\=\dfrac{3}{1}=3$

   Trả lời