1) tính \(\lim\limits_{n\rightarrow\infty}\dfrac{6n-8}{n-1}\)
2) tính \(\lim\limits_{n\rightarrow\infty}\dfrac{n^2+5n-3}{4n^3-2n+5}\)
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MH
13 tháng 10 2023
1) \(\lim\limits_{n\rightarrow\infty}\dfrac{6n-8}{n-1}=\lim\limits_{n\rightarrow\infty}\dfrac{2n\left(1-\dfrac{4}{n}\right)}{n\left(1-\dfrac{1}{n}\right)}=2\)
2) \(\lim\limits_{n\rightarrow\infty}\dfrac{n^2+5n-3}{4n^3-2n+5}=\lim\limits_{n\rightarrow\infty}\dfrac{n^2\left(1+\dfrac{5}{n}-\dfrac{3}{n^2}\right)}{n^3\left(4-\dfrac{2}{n^2}+\dfrac{5}{n^3}\right)}=\dfrac{1}{4n}=\infty\)
3) \(\lim\limits_{n\rightarrow\infty}\left(-2n^5+4n^4-3n^2+4\right)=\lim\limits_{n\rightarrow\infty}n^5\left(-2+\dfrac{4}{n}-\dfrac{3}{n^2}+\dfrac{4}{n^5}\right)=-2n^5=-\infty\)
Y
15 tháng 10 2023
\(1,\lim\limits_{n\rightarrow\infty}\dfrac{-n^2+2n+1}{\sqrt{3n^4+2}}\left(1\right)\)
\(\dfrac{-n^2+2n+1}{\sqrt{3n^4+2}}=\dfrac{-\dfrac{n^2}{n^4}+\dfrac{2n}{n^4}+\dfrac{1}{n^4}}{\sqrt{\dfrac{3n^4}{n^4}+\dfrac{2}{n^4}}}=\dfrac{-\dfrac{1}{n^2}+\dfrac{2}{n^3}+\dfrac{1}{n^4}}{\sqrt{3+\dfrac{2}{n^4}}}\)
\(\Rightarrow\left(1\right)=\dfrac{-lim\dfrac{1}{n^2}+2lim\dfrac{1}{n^3}+lim\dfrac{1}{n^4}}{\sqrt{lim\left(3+\dfrac{2}{n^4}\right)}}\)
\(=\dfrac{0}{\sqrt{lim\left(3+\dfrac{2}{n^4}\right)}}=0\)
\(2,\lim\limits_{n\rightarrow\infty}\left(\dfrac{4n-\sqrt{16n^2+1}}{n+1}\right)\left(2\right)\)
\(\dfrac{4n-\sqrt{16n^2+1}}{n+1}=\dfrac{\dfrac{4n}{n^2}-\sqrt{\dfrac{16n^2}{n^2}+\dfrac{1}{n^2}}}{\dfrac{n}{n^2}+\dfrac{1}{n^2}}=\dfrac{\dfrac{4}{n}-\sqrt{16+\dfrac{1}{n^2}}}{\dfrac{1}{n}+\dfrac{1}{n^2}}\)
\(\Rightarrow\left(2\right)=\dfrac{lim\left(\dfrac{4}{n}-\sqrt{16+\dfrac{1}{n^2}}\right)}{lim\left(\dfrac{1}{n}+\dfrac{1}{n^2}\right)}=\dfrac{lim\left(\dfrac{4}{n}-\sqrt{16+\dfrac{1}{n^2}}\right)}{0}\)
Vậy giới hạn \(\left(2\right)\) không xác định.
\(3,\lim\limits_{n\rightarrow\infty}\left(\dfrac{\sqrt{9n^2+n+1}-3n}{2n}\right)\left(3\right)\)
\(\dfrac{\sqrt{9n^2+n+1}-3n}{2n}=\dfrac{\sqrt{9+\dfrac{1}{n}+\dfrac{1}{n^2}}-\dfrac{3}{n}}{\dfrac{2}{n}}\)
\(\Rightarrow\left(3\right)=\dfrac{lim\left(\sqrt{9+\dfrac{1}{n}+\dfrac{1}{n^2}}-\dfrac{3}{n}\right)}{2lim\dfrac{1}{n}}=\dfrac{lim\left(\sqrt{9+\dfrac{1}{n}+\dfrac{1}{n^2}}-\dfrac{3}{n}\right)}{0}\)
Vậy \(lim\left(3\right)\) không xác định.
24 tháng 11 2023
1: \(\lim\limits_{n\rightarrow\infty}\dfrac{3n^2+5n-3}{-n+5}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2\left(3+\dfrac{5}{n}-\dfrac{3}{n^2}\right)}{n\left(-1+\dfrac{5}{n}\right)}\)
\(=\lim\limits_{n\rightarrow\infty}\left[n\left(\dfrac{3+\dfrac{5}{n}-\dfrac{3}{n^2}}{-1+\dfrac{5}{n}}\right)\right]\)
\(=-\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{n\rightarrow\infty}n=+\infty\\\lim\limits_{n\rightarrow\infty}\dfrac{3+\dfrac{5}{n}-\dfrac{3}{n^2}}{-1+\dfrac{5}{n}}=\dfrac{3+0-0}{-1+0}=\dfrac{3}{-1}=-3< 0\end{matrix}\right.\)
2: \(\lim\limits_{n\rightarrow\infty}\dfrac{-7n^2+4}{-n+5}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{7n^2-4}{n-5}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2\left(7-\dfrac{4}{n^2}\right)}{n\left(1-\dfrac{5}{n}\right)}\)
\(=\lim\limits_{n\rightarrow\infty}\left[n\cdot\dfrac{\left(7-\dfrac{4}{n^2}\right)}{1-\dfrac{5}{n}}\right]\)
\(=+\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{n\rightarrow\infty}n=+\infty\\\lim\limits_{n\rightarrow\infty}\dfrac{7-\dfrac{4}{n^2}}{1-\dfrac{5}{n}}=\dfrac{7-0}{1-0}=7>0\end{matrix}\right.\)
15 tháng 10 2023
1:
\(\lim\limits_{n\rightarrow\infty}\dfrac{3n^5+3n^3-1}{n^3-2n}=\lim\limits_{n\rightarrow\infty}\dfrac{n^5\left(3+\dfrac{3}{n^2}-\dfrac{1}{n^5}\right)}{n^3\left(1-\dfrac{2}{n^2}\right)}\)
\(=\lim\limits_{n\rightarrow\infty}n^2\cdot3=+\infty\)
2: \(\lim\limits_{n\rightarrow\infty}\dfrac{3n^7+3n^5-n}{3n^2-2n}=\lim\limits_{n\rightarrow\infty}\dfrac{3n^6+3n^4-1}{3n-2}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^6\left(3+\dfrac{3}{n^2}-\dfrac{1}{n^6}\right)}{n\left(3-\dfrac{2}{n}\right)}=\lim\limits_{n\rightarrow\infty}n^5=+\infty\)
15 tháng 10 2023
1: \(\lim\limits_{n->\infty}\dfrac{2n+1}{n+15}=\lim\limits_{n\rightarrow\infty}\dfrac{2+\dfrac{1}{n}}{1+\dfrac{15}{n}}=2\)
2: \(\lim\limits_{n\rightarrow\infty}\dfrac{n+6}{2n-5}=\lim\limits_{n\rightarrow\infty}\dfrac{1+\dfrac{6}{n}}{2-\dfrac{5}{n}}=\dfrac{1}{2}\)
15 tháng 10 2023
1:
\(\lim\limits_{n\rightarrow\infty}\dfrac{-3n^3+3n^2-1}{n^2-2n}=\lim\limits_{n\rightarrow\infty}\dfrac{n^3\left(-3+\dfrac{3}{n}-\dfrac{1}{n^3}\right)}{n^2\left(1-\dfrac{2}{n}\right)}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{-3n^3}{n^2}=\lim\limits_{n\rightarrow\infty}-3n=-\infty\)
2:
\(\lim\limits_{n\rightarrow\infty}\dfrac{3n^2-1}{-2n+3}=\lim\limits_{n\rightarrow\infty}\dfrac{n^2\left(3-\dfrac{1}{n^2}\right)}{n\left(-2+\dfrac{3}{n}\right)}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{-3}{2}n=-\infty\)
14 tháng 10 2023
1) \(\lim\limits_{n\rightarrow\infty}\dfrac{3n^2+5n-3}{-n+5}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n\left(3n+5-\dfrac{3}{n}\right)}{-n\left(1-\dfrac{5}{n}\right)}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{3n+5-\dfrac{3}{n}}{-\left(1-\dfrac{5}{n}\right)}\)
\(=\left[{}\begin{matrix}-\infty\left(n\rightarrow+\infty\right)\\+\infty\left(n\rightarrow-\infty\right)\end{matrix}\right.\)
Bài 2,3 tương tự, bạn tự làm nhé!
15 tháng 10 2023
1: \(\lim\limits_{n\rightarrow\infty}\dfrac{n^2-n+2}{n^3+2n^2-3}=\lim\limits_{n\rightarrow\infty}\dfrac{n^2\left(1-\dfrac{1}{n}+\dfrac{2}{n^2}\right)}{n^3\left(1+\dfrac{2}{n}-\dfrac{3}{n^3}\right)}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{1-\dfrac{1}{n}+\dfrac{2}{n^2}}{n\left(1+\dfrac{2}{n}-\dfrac{3}{n^3}\right)}=\lim\limits_{n\rightarrow\infty}\dfrac{1}{n}=0\)
2:
\(\lim\limits_{n\rightarrow\infty}\dfrac{n+2}{3n^3+n^2-2n}=\lim\limits_{n\rightarrow\infty}\dfrac{n\left(1+\dfrac{2}{n}\right)}{n^3\left(3+\dfrac{1}{n}-\dfrac{2}{n^2}\right)}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{1}{n^2}=0\)
1: \(\lim\limits_{n\rightarrow\infty}\dfrac{6n-8}{n-1}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{6-\dfrac{8}{n}}{1-\dfrac{1}{n}}=\dfrac{6-0}{1-0}\)
\(=\dfrac{6}{1}=6\)
2: \(\lim\limits_{n\rightarrow\infty}\dfrac{n^2+5n-3}{4n^3-2n+5}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2\left(1+\dfrac{5}{n}-\dfrac{3}{n^2}\right)}{n^3\left(4-\dfrac{2}{n^2}+\dfrac{5}{n^3}\right)}\)
\(=\lim\limits_{n\rightarrow\infty}\left(\dfrac{1}{n}\cdot\dfrac{1+\dfrac{5}{n}-\dfrac{3}{n^2}}{\left(4-\dfrac{2}{n^2}+\dfrac{5}{n^3}\right)}\right)\)
=0