Tính giới hạn của dãy số B = l i m n 6 + n + 1 3 - 4 n 4 + 2 n - 1 ( 2 n + 3 ) 2
A. +∞
B. -∞
C. 3
D. -3/4
K
Khách
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Những câu hỏi liên quan
NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 1 2021
\(a=lim\dfrac{\left(\dfrac{2}{6}\right)^n+1-\dfrac{1}{4}\left(\dfrac{4}{6}\right)^n}{\left(\dfrac{3}{6}\right)^n+6}=\dfrac{1}{6}\)
\(b=\lim\dfrac{\left(n+1\right)^2}{3n^2+4}=\lim\dfrac{n^2+2n+1}{3n^2+4}=\lim\dfrac{1+\dfrac{2}{n}+\dfrac{1}{n^2}}{3+\dfrac{4}{n^2}}=\dfrac{1}{3}\)
\(c=\lim\dfrac{n\left(n+1\right)}{2\left(n^2-3\right)}=\lim\dfrac{n^2+n}{2n^2-6}=\lim\dfrac{1+\dfrac{1}{n}}{2-\dfrac{6}{n^2}}=\dfrac{1}{2}\)
\(d=\lim\left[1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\right]=\lim\left[1-\dfrac{1}{n+1}\right]=1\)
\(e=\lim\dfrac{1}{2}\left[1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right]\)
\(=\lim\dfrac{1}{2}\left[1-\dfrac{1}{2n+1}\right]=\dfrac{1}{2}\)
22 tháng 10 2023
1: \(-1< =cosx< =1\)
=>\(-3< =3\cdot cosx< =3\)
=>\(y\in\left[-3;3\right]\)
2:
TXĐ là D=R
3: \(L=\lim\limits\dfrac{-3n^3+n^2}{2n^3+5n-2}\)
\(=\lim\limits\dfrac{-3+\dfrac{1}{n}}{2+\dfrac{5}{n^2}-\dfrac{2}{n^3}}=-\dfrac{3}{2}\)
4:
\(L=lim\left(3n^2+5n-3\right)\)
\(=\lim\limits\left[n^2\left(3+\dfrac{5}{n}-\dfrac{3}{n^2}\right)\right]\)
\(=+\infty\) vì \(\left\{{}\begin{matrix}lim\left(n^2\right)=+\infty\\\lim\limits\left(3+\dfrac{5}{n}-\dfrac{3}{n^2}\right)=3>0\end{matrix}\right.\)
5:
\(\lim\limits_{n\rightarrow+\infty}n^3-2n^2+3n-4\)
\(=\lim\limits_{n\rightarrow+\infty}n^3\left(1-\dfrac{2}{n}+\dfrac{3}{n^2}-\dfrac{4}{n^3}\right)\)
\(=+\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{n\rightarrow+\infty}n^3=+\infty\\\lim\limits_{n\rightarrow+\infty}1-\dfrac{2}{n}+\dfrac{3}{n^2}-\dfrac{4}{n^3}=1>0\end{matrix}\right.\)
TT
22 tháng 10 2023
\(1,y=3cosx\)
\(+TXD\) \(D=R\)
Có \(-1\le cosx\le1\)
\(\Leftrightarrow-3\le3cosx\le3\)
Vậy có tập giá trị \(T=\left[-3;3\right]\)
\(2,y=cosx\)
\(TXD\) \(D=R\)
\(3,L=lim\dfrac{n^2-3n^3}{2n^3+5n-2}=lim\dfrac{\dfrac{1}{n}-3}{2+\dfrac{5}{n^2}-\dfrac{2}{n^3}}\)(chia cả tử và mẫu cho \(n^3\))
\(=\dfrac{lim\dfrac{1}{n}-lim3}{lim2+5lim\dfrac{1}{n^2}-2lim\dfrac{1}{n^3}}=\dfrac{0-3}{2+5.0-2.0}=-\dfrac{3}{2}\)
\(4,L=lim\left(3n^2+5n-3\right)\\ =lim\left(3+\dfrac{5}{n}-\dfrac{3}{n^2}\right)\\ =lim3+5lim\dfrac{1}{n}-3lim\dfrac{1}{n^2}\\ =3\)
\(5,\lim\limits_{n\rightarrow+\infty}\left(n^3-2n^2+3n-4\right)\\ =lim\left(1-\dfrac{2}{n}+\dfrac{3}{n^2}-\dfrac{4}{n^3}\right)\\ =lim1-0\\ =1\)
15 tháng 10 2023
3:
\(\lim\limits_{n\rightarrow\infty}\dfrac{2-5^{n-2}}{3^n+2\cdot5^n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{\dfrac{2}{5^n}-\dfrac{5^{n-2}}{5^n}}{\dfrac{3^n}{5^n}+2\cdot\dfrac{5^n}{5^n}}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{\dfrac{2}{5^n}-\dfrac{1}{25}}{\left(\dfrac{3}{5}\right)^n+2\cdot1}\)
\(=-\dfrac{1}{25}:2=-\dfrac{1}{50}\)
1:
\(=\lim\limits_{n\rightarrow\infty}\dfrac{3^n-4^n\cdot4}{3^n\cdot9+4^n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{\dfrac{3^n}{4^n}-4}{3^n\cdot\dfrac{9}{4^n}+1}\)
\(=-\dfrac{4}{1}=-4\)
HT
18 tháng 2 2021
1/ \(\lim\limits\dfrac{\dfrac{2^n}{7^n}-5.7.\left(\dfrac{7}{7}\right)^n}{\dfrac{2^n}{7^n}+\left(\dfrac{7}{7}\right)^n}=-35\)
2/ \(\lim\limits\dfrac{\dfrac{3^n}{7^n}-2.5.\left(\dfrac{5}{7}\right)^n}{\dfrac{2^n}{7^n}+\dfrac{7^n}{7^n}}=0\)
3/ \(\lim\limits\sqrt[3]{\dfrac{\dfrac{5}{n}-\dfrac{8n}{n}}{\dfrac{n}{n}+\dfrac{3}{n}}}=\sqrt[3]{-8}=-2\)
HT
1 tháng 2 2021
Một câu thôi: Liên hợp
\(\dfrac{1}{2\sqrt{1}+\sqrt{2}}=\dfrac{2.1-\sqrt{2}}{2^2-2}=\dfrac{2-\sqrt{2}}{2}=1-\dfrac{1}{\sqrt{2}}\)
\(\dfrac{1}{3\sqrt{2}+2\sqrt{3}}=\dfrac{3\sqrt{2}-2\sqrt{3}}{9.2-4.3}=\dfrac{3\sqrt{2}-2\sqrt{3}}{6}=\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}\)
\(\Rightarrow\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
Nên chứng minh bằng quy nạp mạnh cho chặt chẽ, giờ tui buồn ngủ quá nên bạn tự chứng minh nha :(
\(\Rightarrow u_n=1-\dfrac{1}{\sqrt{n+1}}=\dfrac{\sqrt{n+1}-1}{\sqrt{n+1}}\Rightarrow\lim\limits\left(u_n\right)=\lim\limits\dfrac{\sqrt{\dfrac{n}{n}+\dfrac{1}{n}}-\dfrac{1}{\sqrt{n}}}{\sqrt{\dfrac{n}{n}+\dfrac{1}{n}}}=1\)
15 tháng 10 2023
2:
\(\lim\limits_{n\rightarrow\infty}\dfrac{3^n+1}{2^n-1}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{\dfrac{3^n}{3^n}+\dfrac{1}{3^n}}{\dfrac{2^n}{3^n}-\dfrac{1}{3^n}}=\lim\limits_{n\rightarrow\infty}\dfrac{1+\dfrac{1}{3^n}}{\left(\dfrac{2}{3}\right)^n-\dfrac{1}{3^n}}=1\)
15 tháng 10 2023
1:
\(K=\lim\limits_{n\rightarrow\infty}\dfrac{3\cdot2^n-3^n}{2^{n+1}+3^{n+1}}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{3\cdot2^n-3^n}{2^n\cdot2+3^n\cdot3}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{3\cdot\dfrac{2^n}{3^n}-1}{\left(\dfrac{2}{3}\right)^n\cdot2+3}\)
\(=-\dfrac{1}{3}\)
2:
\(\lim\limits_{n\rightarrow\infty}\dfrac{3^n-4^{n+1}}{3^{n+2}+4^n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{3^n-4^n\cdot4}{3^n\cdot9+4^n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{\left(\dfrac{3}{4}\right)^n-4}{\left(\dfrac{3}{4}\right)^n\cdot9+1}=-\dfrac{4}{1}=-4\)
Chọn D.
Chia cả tử và mẫu cho n2 ta có được: