Tính giới hạn: l i m 1 - 1 2 2 1 - 1 3 2 . . . . . . 1 - 1 n 2
A. 1.
B. 1/2.
C. 2.
D. 3.
K
Khách
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AH
Akai Haruma
Giáo viên
27 tháng 2 2022
Lời giải:
\(\lim\frac{6n^3-2n+1}{(5n^3-n)(n^2+n-1)}=\lim \frac{6-\frac{2}{n^2}+\frac{1}{n^3}}{(5-\frac{1}{n^2})(n^2+n-1)}\)
Ta thấy:
\(\lim\frac{6-\frac{2}{n^2}+\frac{1}{n^3}}{5-\frac{1}{n^2}}=\frac{6}{5}\)
\(\lim \frac{1}{n^2+n-1}=0\)
$\Rightarrow L=0$
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.
15 tháng 10 2023
1:
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2-1-9n^2}{\sqrt{n^2-1}-3n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{-8n^2-1}{\sqrt{n^2-1}-3n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2\left(-8-\dfrac{1}{n^2}\right)}{n\left(\sqrt{1-\dfrac{1}{n^2}}-3\right)}=\lim\limits_{n\rightarrow\infty}-\dfrac{8}{1-3}\cdot n=\lim\limits_{n\rightarrow\infty}4n=+\infty\)
2:
\(\lim\limits_{n\rightarrow\infty}\sqrt{4n^2+5}+n\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{4n^2+5-n^2}{\sqrt{4n^2+5}-n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{3n^2+5}{\sqrt{4n^2+5}-n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2\left(3+\dfrac{5}{n^2}\right)}{n\left(\sqrt{4+\dfrac{5}{n^2}}-1\right)}\)
\(=\lim\limits_{n\rightarrow\infty}n\cdot\left(\dfrac{3}{\sqrt{4}-1}\right)=+\infty\)
15 tháng 10 2023
2:
\(\lim\limits_{n\rightarrow\infty}\dfrac{3^n+1}{2^n-1}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{\dfrac{3^n}{3^n}+\dfrac{1}{3^n}}{\dfrac{2^n}{3^n}-\dfrac{1}{3^n}}=\lim\limits_{n\rightarrow\infty}\dfrac{1+\dfrac{1}{3^n}}{\left(\dfrac{2}{3}\right)^n-\dfrac{1}{3^n}}=1\)
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 \(\left\{{}\begin{matrix}\lim\limits_{n\rightarrow+\infty}n^3=+\infty\\\lim\limits_{n\rightarrow+\infty}1-\dfrac{2}{n}+\dfrac{3}{n^2}-\dfrac{4}{n^3}=1>0\end{matrix}\right.\)
Y
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
1:
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^3+3n^2+1-n^3}{\sqrt[3]{n^3+3n^2+1}+n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{3n^2+1}{\sqrt[3]{n^3+3n^2+1}+n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2\left(3+\dfrac{1}{n^2}\right)}{n\left(\sqrt[3]{1+\dfrac{3}{n}+\dfrac{1}{n^3}}+1\right)}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n\cdot\left(3+\dfrac{1}{n^2}\right)}{\sqrt[3]{1+\dfrac{3}{n}+\dfrac{1}{n^3}}+1}\)
\(=\lim\limits_{n\rightarrow\infty}n\cdot\lim\limits_{n\rightarrow\infty}\dfrac{3+\dfrac{1}{n^2}}{\sqrt[3]{1+\dfrac{3}{n}+\dfrac{1}{n^3}}+1}\)
\(=+\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{n\rightarrow\infty}n=+\infty\\\lim\limits_{n\rightarrow\infty}\dfrac{3+\dfrac{1}{n^2}}{\sqrt[3]{1+\dfrac{3}{n}+\dfrac{1}{n^3}}+1}=\dfrac{3}{2}>0\end{matrix}\right.\)
2:
\(=\lim\limits_{n\rightarrow\infty}\left(\sqrt{4n^2+1}-2n+2n-\sqrt[3]{8n^3+n}\right)\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{4n^2+1-4n^2}{\sqrt{4n^2+1}+2n}+\lim\limits_{n\rightarrow\infty}\dfrac{8n^3-8n^3-n}{4n^2+2n\cdot\sqrt[3]{8n^3+n}+\left(\sqrt[3]{8n^3+n}\right)^2}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{1}{\sqrt{4n^2+1}+2n}+\lim\limits_{n\rightarrow\infty}\dfrac{-n}{4n^2+2n\cdot n\cdot\sqrt[3]{8+\dfrac{1}{n^3}}+\left(n\cdot\sqrt[3]{8+\dfrac{1}{n^3}}\right)^2}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{-n}{4n^2+2n^2\cdot\sqrt[3]{8+\dfrac{1}{n^3}}+n^2\cdot\sqrt[3]{\left(8+\dfrac{1}{n^3}\right)^2}}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{-1}{4n+2n\cdot\sqrt[3]{8+\dfrac{1}{n^3}}+n\cdot\sqrt[3]{\left(8+\dfrac{1}{n^3}\right)^2}}\)
\(=0\)
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:
\(=\lim\limits_{n\rightarrow\infty}\dfrac{3^n-4^n\cdot4}{3^n\cdot9+4^n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{\dfrac{3^n}{4^n}-4}{3^n\cdot\dfrac{9}{4^n}+1}\)
\(=-\dfrac{4}{1}=-4\)
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:
\(\lim\limits_{n\rightarrow\infty}\dfrac{3^n-4^{n+1}}{3^{n+2}+4^n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{3^n-4^n\cdot4}{3^n\cdot9+4^n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{\left(\dfrac{3}{4}\right)^n-4}{\left(\dfrac{3}{4}\right)^n\cdot9+1}=-\dfrac{4}{1}=-4\)
AH
Akai Haruma
Giáo viên
3 tháng 4 2022
Lời giải:
\(L=\lim\limits_{x\to 1}\frac{\sqrt{2x-1}(\sqrt[3]{x+7}-2)+2(\sqrt{2x-1}-1)}{x(x-1)}=\lim\limits_{x\to 1}\frac{\sqrt{2x-1}.\frac{1}{\sqrt[3]{(x+7)^2}+2\sqrt[3]{x+7}+4}+4.\frac{1}{\sqrt{2x-1}+1}}{x}=\frac{25}{12}\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 1 2021
\(=\lim\dfrac{1+\dfrac{1}{n}+\dfrac{5}{n^2}}{2+\dfrac{1}{n^2}}=\dfrac{1}{2}\)
Chọn B.
Ta có: