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NV
Nguyễn Việt Lâm
Giáo viên
2 tháng 1 2019
\(lim\left(5n-\sqrt{25n^2-3n+5}\right)=lim\dfrac{25n^2-25n^2+3n-5}{5n+\sqrt{25n^2-3n+5}}\)
\(=lim\dfrac{3n-5}{5n+\sqrt{25n^2-3n+5}}=lim\dfrac{3-\dfrac{5}{n}}{5+\sqrt{25-\dfrac{3}{n}+\dfrac{5}{n^2}}}=\dfrac{3-0}{5+\sqrt{25-0+0}}=\dfrac{3}{10}\)
\(lim\dfrac{4n^5-3n^4-2n^3+7n-9}{-5n\left(3n^2-3n+1\right)\left(5-2n^2\right)}=lim\dfrac{\dfrac{4n^5-3n^4-2n^3+7n-9}{n^5}}{\dfrac{-5n}{n}\dfrac{\left(3n^2-3n+1\right)}{n^2}\dfrac{\left(5-2n^2\right)}{n^2}}\)
\(=lim\dfrac{4-\dfrac{3}{n}-\dfrac{2}{n^2}+\dfrac{7}{n^4}-\dfrac{9}{n^5}}{-5.\left(3-\dfrac{2}{n}+\dfrac{1}{n^2}\right).\left(\dfrac{5}{n^2}-2\right)}=\dfrac{4-0-0+0-0}{-5\left(3-0+0\right).\left(0-2\right)}=\dfrac{2}{15}\)
HM
1
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 1 2019
\(lim\dfrac{5n\sqrt{2n^2-n}}{1+5n-3n^2}=lim\dfrac{5\sqrt{2-\dfrac{1}{n}}}{\dfrac{1}{n^2}+\dfrac{5}{n}-3}=\dfrac{5\sqrt{2-0}}{0+0-3}=\dfrac{-5\sqrt{2}}{3}\)
\(lim\dfrac{\sqrt{4n^2+n}-7n}{3n^2-1}=lim\dfrac{\sqrt{\dfrac{4}{n^2}+\dfrac{1}{n^3}}-\dfrac{7}{n}}{3-\dfrac{1}{n^2}}=\dfrac{\sqrt{0+0}-0}{3-0}=\dfrac{0}{3}=0\)
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 1 2021
\(a=\lim\left(\dfrac{2n^3\left(5n+1\right)+\left(2n^2+3\right)\left(1-5n^2\right)}{\left(2n^2+3\right)\left(5n+1\right)}\right)\)
\(=\lim\left(\dfrac{2n^3-13n^2+3}{\left(2n^2+3\right)\left(5n+1\right)}\right)=\lim\dfrac{2-\dfrac{13}{n}+\dfrac{3}{n^3}}{\left(2+\dfrac{3}{n^2}\right)\left(5+\dfrac{1}{n}\right)}=\dfrac{2}{2.5}=\dfrac{1}{5}\)
\(b=\lim\left(\dfrac{n-2}{\sqrt{n^2+n}+\sqrt{n^2+2}}\right)=\lim\dfrac{1-\dfrac{2}{n}}{\sqrt{1+\dfrac{1}{n}}+\sqrt{1+\dfrac{2}{n}}}=\dfrac{1}{2}\)
\(c=\lim\dfrac{\sqrt{1+\dfrac{3}{n^3}-\dfrac{2}{n^4}}}{2-\dfrac{2}{n}+\dfrac{3}{n^2}}=\dfrac{1}{2}\)
\(d=\lim\dfrac{\sqrt{1-\dfrac{4}{n}}-\sqrt{4+\dfrac{1}{n^2}}}{\sqrt{3+\dfrac{1}{n^2}}-1}=\dfrac{1-2}{\sqrt{3}-1}=-\dfrac{1+\sqrt{3}}{2}\)
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!
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 \(\left\{{}\begin{matrix}\lim\limits_{n\rightarrow\infty}n=+\infty\\\lim\limits_{n\rightarrow\infty}\dfrac{7-\dfrac{4}{n^2}}{1-\dfrac{5}{n}}=\dfrac{7-0}{1-0}=7>0\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
29 tháng 1 2023
a.
\(A=\lim\frac{\sqrt[3]{n^6-7n^3-5n+8}}{n+12}=\lim \frac{\sqrt[3]{\frac{n^6-7n^3-5n+8}{n^3}}}{\frac{n+12}{n}}=\lim \frac{\sqrt[3]{n^3-7-\frac{5}{n^2}+\frac{8}{n^3}}}{1+\frac{12}{n}}\)
Ta thấy:
\(\lim\sqrt[3]{n^3-7-\frac{5}{n^2}+\frac{8}{n^3}}=\infty \)
\(\lim (1+\frac{12}{n})=1\)
Suy ra $A=\infty$
AH
Akai Haruma
Giáo viên
29 tháng 1 2023
b.
\(B=\lim\frac{1}{\sqrt{3n+2}-\sqrt{2n+1}}=\lim \frac{1}{\frac{3n+2-(2n+1)}{\sqrt{3n+2}+\sqrt{2n+1}}}=\lim \frac{\sqrt{3n+2}+\sqrt{2n+1}}{n+1}\)
\(=\lim \frac{\sqrt{\frac{3n+2}{n}}+\sqrt{\frac{2n+1}{n}}}{\frac{n+1}{\sqrt{n}}}=\lim \frac{\sqrt{3+\frac{2}{n}}+\sqrt{2+\frac{1}{n}}}{\sqrt{n}+\frac{1}{\sqrt{n}}}\)
Ta thấy:
\(\lim( \sqrt{3+\frac{2}{n}}+\sqrt{2+\frac{1}{n}})=\sqrt{3}+\sqrt{2}>0\)
\(\lim (\sqrt{n}+\frac{1}{\sqrt{n}})=\infty\)
$\Rightarrow B=\infty$
AH
Akai Haruma
Giáo viên
10 tháng 4 2020
1.
\(\lim \frac{3n^2+5n+4}{2-n^2}=\lim \frac{\frac{3n^2+5n+4}{n^2}}{\frac{2-n^2}{n^2}}=\lim \frac{3+\frac{5}{n}+\frac{4}{n^2}}{\frac{2}{n^2}-1}=\frac{3}{-1}=-3\)
2.
\(\lim \frac{2n^3-4n^2+3n+7}{n^3-7n+5}=\lim \frac{\frac{2n^3-4n^2+3n+7}{n^3}}{\frac{n^3-7n+5}{n^3}}=\lim \frac{2-\frac{4}{n}+\frac{3}{n^2}+\frac{7}{n^3}}{1-\frac{7}{n^2}+\frac{5}{n^3}}=\frac{2}{1}=2\)
3.
\(\lim (\frac{2n^3}{2n^2+3}+\frac{1-5n^2}{5n+1})=\lim (n-\frac{3n}{2n^2+3}+\frac{1}{5}-n-\frac{1}{5n+1})\)
\(=\frac{1}{5}-\lim (\frac{3n}{2n^2+3}+\frac{1}{5n+1})=\frac{1}{5}-\lim (\frac{3}{2n+\frac{3}{n}}+\frac{1}{5n+1})=\frac{1}{5}-0=\frac{1}{5}\)
4.
\(\lim \frac{1+3^n}{4+3^n}=\lim (1-\frac{3}{4+3^n})=1-\lim \frac{3}{4+3^n}=1-0=1\)
5.
\(\lim \frac{4.3^n+7^{n+1}}{2.5^n+7^n}=\lim \frac{\frac{4.3^n+7^{n+1}}{7^n}}{\frac{2.5^n+7^n}{7^n}}\)
\(=\lim \frac{4.(\frac{3}{7})^n+7}{2.(\frac{5}{7})^n+1}=\frac{7}{1}=7\)
DG
4
MH
11 tháng 2 2022
\(b,lim\dfrac{2n^2+1}{3n^3-3n+3}\)
\(=lim\dfrac{2n+\dfrac{1}{n^3}}{3-\dfrac{3}{n^2}+\dfrac{3}{n^3}}\)
\(=n\times\dfrac{2}{3}=\)+∞
lim= \(\dfrac{n^3\left(5-\dfrac{3}{n}+\dfrac{6}{n^3}\right)}{n^3\left(\dfrac{4}{n}-3+\dfrac{7}{n^2}\right)}\)
lim= \(\dfrac{5}{-3}\)