1. Tìm lim (1 – n^2) / (2n^2 + 1) 2. Tìm lim (4n + 2018) / (2n + 1) 05/11/2021 Bởi Lydia 1. Tìm lim (1 – n^2) / (2n^2 + 1) 2. Tìm lim (4n + 2018) / (2n + 1)
Đáp án: $\begin{array}{l}1)\lim \dfrac{{1 – {n^2}}}{{2{n^2} + 1}} = \lim \dfrac{{\dfrac{1}{{{n^2}}} – 1}}{{2 + \dfrac{1}{{{n^2}}}}} = \dfrac{{ – 1}}{2}\\2)\lim \dfrac{{4n + 2018}}{{2n + 1}} = \lim \dfrac{{4 + \dfrac{{2018}}{n}}}{{2 + \dfrac{1}{n}}} = \dfrac{4}{2} = 2\end{array}$ Bình luận
1. $\lim\dfrac{1-n^2}{2n^2+1}$ $=\lim\dfrac{\dfrac{1}{n^2}-1}{2+\dfrac{1}{n^2}}$ $=\dfrac{-1}{2}$ 2. $\lim\dfrac{4n+2018}{2n+1}$ $=\lim\dfrac{4+\dfrac{2018}{n}}{2+\dfrac{1}{n}}$ $=\dfrac{4}{2}=2$ Bình luận
Đáp án:
$\begin{array}{l}
1)\lim \dfrac{{1 – {n^2}}}{{2{n^2} + 1}} = \lim \dfrac{{\dfrac{1}{{{n^2}}} – 1}}{{2 + \dfrac{1}{{{n^2}}}}} = \dfrac{{ – 1}}{2}\\
2)\lim \dfrac{{4n + 2018}}{{2n + 1}} = \lim \dfrac{{4 + \dfrac{{2018}}{n}}}{{2 + \dfrac{1}{n}}} = \dfrac{4}{2} = 2
\end{array}$
1.
$\lim\dfrac{1-n^2}{2n^2+1}$
$=\lim\dfrac{\dfrac{1}{n^2}-1}{2+\dfrac{1}{n^2}}$
$=\dfrac{-1}{2}$
2.
$\lim\dfrac{4n+2018}{2n+1}$
$=\lim\dfrac{4+\dfrac{2018}{n}}{2+\dfrac{1}{n}}$
$=\dfrac{4}{2}=2$