1. Tìm lim (3n^2 + 1) / (n^2 – 2) 2. Tìm lim (8n^2 + 3n – 1) / (4 + 5n + 2n^2) 05/11/2021 Bởi Amara 1. Tìm lim (3n^2 + 1) / (n^2 – 2) 2. Tìm lim (8n^2 + 3n – 1) / (4 + 5n + 2n^2)
1) $\,\,\,\,\lim \dfrac{3{{n}^{2}}+1}{{{n}^{2}}-2}$ $=\lim \dfrac{3+\dfrac{1}{{{n}^{2}}}}{1-\dfrac{2}{{{n}^{2}}}}$ $=\dfrac{3}{1}$ $=3$ 2) $\lim \dfrac{8{{n}^{2}}+3n-1}{4+5n+2{{n}^{2}}}$ $=\lim \dfrac{8{{n}^{2}}+3n-1}{2{{n}^{2}}+5n+4}$ $=\lim \dfrac{8+\frac{3}{n}-\dfrac{1}{{{n}^{2}}}}{2+\dfrac{5}{n}+\dfrac{4}{{{n}^{2}}}}$ $=\dfrac{8}{2}$ $=4$ Bình luận
1. $\lim\dfrac{3n^2+1}{n^2-2}$ $=\lim\dfrac{3+\dfrac{1}{n^2}}{1-\dfrac{2}{n^2}}$ $=\dfrac{3}{1}=3$ 2. $\lim\dfrac{8n^2+3n-1}{2n^2+5n+4}$ $=\lim\dfrac{8+\dfrac{3}{n}-\dfrac{1}{n^2}}{2+\dfrac{5}{n}+\dfrac{4}{n^2}}$ $=\dfrac{8}{2}=4$ Bình luận
1)
$\,\,\,\,\lim \dfrac{3{{n}^{2}}+1}{{{n}^{2}}-2}$
$=\lim \dfrac{3+\dfrac{1}{{{n}^{2}}}}{1-\dfrac{2}{{{n}^{2}}}}$
$=\dfrac{3}{1}$
$=3$
2)
$\lim \dfrac{8{{n}^{2}}+3n-1}{4+5n+2{{n}^{2}}}$
$=\lim \dfrac{8{{n}^{2}}+3n-1}{2{{n}^{2}}+5n+4}$
$=\lim \dfrac{8+\frac{3}{n}-\dfrac{1}{{{n}^{2}}}}{2+\dfrac{5}{n}+\dfrac{4}{{{n}^{2}}}}$
$=\dfrac{8}{2}$
$=4$
1.
$\lim\dfrac{3n^2+1}{n^2-2}$
$=\lim\dfrac{3+\dfrac{1}{n^2}}{1-\dfrac{2}{n^2}}$
$=\dfrac{3}{1}=3$
2.
$\lim\dfrac{8n^2+3n-1}{2n^2+5n+4}$
$=\lim\dfrac{8+\dfrac{3}{n}-\dfrac{1}{n^2}}{2+\dfrac{5}{n}+\dfrac{4}{n^2}}$
$=\dfrac{8}{2}=4$