PT
1
K
Khách
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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:
\(\lim\limits_{n\rightarrow\infty}\sqrt{4n^2+5}+n\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{4n^2+5-n^2}{\sqrt{4n^2+5}-n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{3n^2+5}{\sqrt{4n^2+5}-n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2\left(3+\dfrac{5}{n^2}\right)}{n\left(\sqrt{4+\dfrac{5}{n^2}}-1\right)}\)
\(=\lim\limits_{n\rightarrow\infty}n\cdot\left(\dfrac{3}{\sqrt{4}-1}\right)=+\infty\)
20 tháng 11 2023
\(\lim\limits\dfrac{\sqrt{4n^2+1}+2n-1}{\sqrt{n^2+4n+1}+n}\)
\(=\lim\limits\dfrac{\sqrt{4+\dfrac{1}{n^2}}+2-\dfrac{1}{n}}{\sqrt{1+\dfrac{4}{n}+\dfrac{1}{n^2}}+1}=\dfrac{2+2}{1+1}=\dfrac{4}{2}=2\)
\(\lim\limits\left[\sqrt{n}\left(\sqrt{n+1}-n\right)\right]\)
\(=\lim\limits\left[\sqrt{n^2+n}-\sqrt{n^3}\right]\)
\(=\lim\limits\dfrac{n^2+n-n^3}{\sqrt{n^2+n}+\sqrt{n^3}}\)
\(=\lim\limits\dfrac{n^3\left(-1+\dfrac{1}{n}+\dfrac{1}{n^2}\right)}{\sqrt{n^3\left(\dfrac{1}{n}+\dfrac{1}{n^2}\right)}+\sqrt{n^3}}\)
\(=\lim\limits\dfrac{n^3\left(-1+\dfrac{1}{n}+\dfrac{1}{n^2}\right)}{\sqrt{n^3}\left(\sqrt{\dfrac{1}{n}+\dfrac{1}{n^2}}+1\right)}\)
\(=\lim\limits\dfrac{n\sqrt{n}\left(-1+\dfrac{1}{n}+\dfrac{1}{n^2}\right)}{\sqrt{\dfrac{1}{n}+\dfrac{1}{n^2}}+1}\)
\(=-\infty\) vì \(\left\{{}\begin{matrix}lim\left(n\sqrt{n}\right)=+\infty\\lim\left(\dfrac{-1+\dfrac{1}{n}+\dfrac{1}{n^2}}{\sqrt{\dfrac{1}{n}+\dfrac{1}{n^2}}+1}\right)=-\dfrac{1}{1}=-1< 0\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 1 2021
\(a=\lim\dfrac{\left(n-2\right)!\left(n-1+\left(n-1\right)n\right)}{\left(n-2\right)!\left(\left(n+2\right)\left(n+1\right)n\left(n-1\right)-1\right)}+\lim\dfrac{3}{\left(n+2\right)!-\left(n-2\right)!}\)
\(=\lim\dfrac{n^2-1}{\left(n+2\right)\left(n+1\right)n\left(n-1\right)-1}+\lim\dfrac{3}{\left(n+2\right)!-\left(n-2\right)!}\)
\(=0+0=0\)
\(b=\lim\dfrac{2+\dfrac{1}{n}}{3^n}=\dfrac{2}{\infty}=0\)
PT
1
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 1 2021
\(=\lim\dfrac{1+\dfrac{1}{n}+\dfrac{5}{n^2}}{2+\dfrac{1}{n^2}}=\dfrac{1}{2}\)
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.
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:
\(\lim\frac{6-\frac{2}{n^2}+\frac{1}{n^3}}{5-\frac{1}{n^2}}=\frac{6}{5}\)
\(\lim \frac{1}{n^2+n-1}=0\)
$\Rightarrow L=0$
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:
\(=\lim\limits_{n\rightarrow\infty}\dfrac{3^n-4^n\cdot4}{3^n\cdot9+4^n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{\dfrac{3^n}{4^n}-4}{3^n\cdot\dfrac{9}{4^n}+1}\)
\(=-\dfrac{4}{1}=-4\)
R
Chứng minh các giới hạn sau :
\(1.lim\dfrac{n^2+1}{n}=+\infty\)
\(2.lim\dfrac{2-n}{\sqrt{n}}=-\infty\)
1
AM
10 tháng 2 2022
1. \(lim_{n\rightarrow+\infty}\dfrac{n^2+1}{n}=lim_{n\rightarrow+\infty}\left(n+\dfrac{1}{n}\right)=+\infty\)(đpcm)
2. \(lim_{n\rightarrow+\infty}\dfrac{2-n}{\sqrt{n}}=lim_{n\rightarrow+\infty}\left(2-\sqrt{n}\right)=-\infty\) (đpcm)
15 tháng 10 2023
1: \(I=\lim\limits_{n\rightarrow\infty}\dfrac{n^2+2-n^2+1}{\sqrt{n^2+2}+\sqrt{n^2-1}}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{3}{\sqrt{n^2+2}+\sqrt{n^2-1}}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{3}{n\left(\sqrt{1+\dfrac{2}{n^2}}+\sqrt{1-\dfrac{1}{n^2}}\right)}\)
=0
2: \(\lim\limits_{n\rightarrow\infty}\sqrt{n^2+2n+2}+n\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2+2n+2-n^2}{\sqrt{n^2+2n+2}-n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{2n+2}{\sqrt{n^2+2n+2}-n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{2+\dfrac{1}{n}}{\sqrt{1+\dfrac{2}{n}+\dfrac{2}{n^2}}-1}\)
\(=+\infty\)
l i m n 2 + 2 n + 1 - n 2 + n - 1 = 1 2