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DG
1
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 1 2021
\(a=\lim\left(\dfrac{2n^3\left(5n+1\right)+\left(2n^2+3\right)\left(1-5n^2\right)}{\left(2n^2+3\right)\left(5n+1\right)}\right)\)
\(=\lim\left(\dfrac{2n^3-13n^2+3}{\left(2n^2+3\right)\left(5n+1\right)}\right)=\lim\dfrac{2-\dfrac{13}{n}+\dfrac{3}{n^3}}{\left(2+\dfrac{3}{n^2}\right)\left(5+\dfrac{1}{n}\right)}=\dfrac{2}{2.5}=\dfrac{1}{5}\)
\(b=\lim\left(\dfrac{n-2}{\sqrt{n^2+n}+\sqrt{n^2+2}}\right)=\lim\dfrac{1-\dfrac{2}{n}}{\sqrt{1+\dfrac{1}{n}}+\sqrt{1+\dfrac{2}{n}}}=\dfrac{1}{2}\)
\(c=\lim\dfrac{\sqrt{1+\dfrac{3}{n^3}-\dfrac{2}{n^4}}}{2-\dfrac{2}{n}+\dfrac{3}{n^2}}=\dfrac{1}{2}\)
\(d=\lim\dfrac{\sqrt{1-\dfrac{4}{n}}-\sqrt{4+\dfrac{1}{n^2}}}{\sqrt{3+\dfrac{1}{n^2}}-1}=\dfrac{1-2}{\sqrt{3}-1}=-\dfrac{1+\sqrt{3}}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
7 tháng 2 2021
\(a=\lim\dfrac{5n\left(n+\sqrt{n^2-n-1}\right)}{n+1}=\lim\dfrac{5\left(n+\sqrt{n^2-n-1}\right)}{1+\dfrac{1}{n}}=\dfrac{+\infty}{1}=+\infty\)
\(b=\lim\dfrac{\sqrt{\dfrac{1}{n}+\sqrt{\dfrac{1}{n^3}+\dfrac{1}{n^4}}}}{1-\dfrac{1}{\sqrt{n}}}=\dfrac{0}{1}=0\)
\(c=\lim\dfrac{\sqrt{2n^2-1+\dfrac{7}{n^2}}}{3+\dfrac{5}{n}}=\dfrac{+\infty}{3}=+\infty\)
\(d=\lim\dfrac{\sqrt{3+\dfrac{2}{n}}-1}{3-\dfrac{2}{n}}=\dfrac{\sqrt{3}-1}{3}\)
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.
\(\lim \frac{4.3^n+7^{n+1}}{2.5^n+7^n}=\lim \frac{\frac{4.3^n+7^{n+1}}{7^n}}{\frac{2.5^n+7^n}{7^n}}\)
\(=\lim \frac{4.(\frac{3}{7})^n+7}{2.(\frac{5}{7})^n+1}=\frac{7}{1}=7\)
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) \(\lim\limits_{n\rightarrow\infty}\left(-2n^5+4n^4-3n^2+4\right)=\lim\limits_{n\rightarrow\infty}n^5\left(-2+\dfrac{4}{n}-\dfrac{3}{n^2}+\dfrac{4}{n^5}\right)=-2n^5=-\infty\)
lim \(\frac{-3n^2+5n+1}{2n^2-n+3}\)
= lim \(\frac{-3+\frac{5}{n}+\frac{1}{n^2}}{2-\frac{1}{n}+\frac{3}{n^2}}\)
= -3/2