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NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 1 2019
\(lim\dfrac{\left(n+2\right)^{50}\left(n-3\right)^{80}}{\left(2n-1\right)^{40}\left(3n-2\right)^{45}}=lim\dfrac{\left(1+\dfrac{2}{n^{50}}\right)\left(1-\dfrac{3}{n^{35}}\right)\left(n-3\right)^{45}}{\left(2-\dfrac{1}{n^{50}}\right)\left(3-\dfrac{2}{n^{45}}\right)}=+\infty\)
\(lim\dfrac{4^n}{2.3^n+4^n}=lim\dfrac{1}{2.\left(\dfrac{3}{4}\right)^n+1}=\dfrac{1}{0+1}=1\)
\(lim\dfrac{3^n-2.5^n}{7+3.5^n}=lim\dfrac{\left(\dfrac{3}{5}\right)^n-2}{\dfrac{7}{5^n}+3}=\dfrac{0-2}{0+3}=\dfrac{-2}{3}\)
\(lim\dfrac{4^n-5^n}{2^{2n}+3.5^{2n}}=lim\dfrac{\left(\dfrac{4}{25}\right)^n-\left(\dfrac{1}{5}\right)^n}{\left(\dfrac{2}{5}\right)^{2n}+3}=\dfrac{0-0}{0+3}=0\)
\(lim\dfrac{\left(-3\right)^n+5^n}{2.\left(-4\right)^n+5^n}=lim\dfrac{\left(\dfrac{-3}{5}\right)^n+1}{2.\left(-\dfrac{4}{5}\right)^n+1}=\dfrac{0+1}{0+1}=1\)
AH
Akai Haruma
Giáo viên
12 tháng 1 2019
1.
Nhớ rằng \(\lim _{x\to \infty}\frac{1}{x}=0\) và \(\lim _{x\to a}\frac{f(x)}{g(x)}=\frac{\lim_{x\to a}f(x)}{\lim_{x\to a}g(x)}\) với \(g(x)\neq 0; \lim_{x\to a}g(x)\neq 0\)
Do đó:
\(\lim_{n\to \infty}\frac{(n+2)^{50}.(n-3)^{80}}{(2n-1)^{40}.(3n-2)^{45}}=\lim_{n\to \infty}\frac{n^{130}(\frac{n+2}{n})^{50}.(\frac{n-3}{n})^{80}}{n^{85}(\frac{2n-1}{n})^{40}.(\frac{3n-2}{n})^{45}}\)
\(=\lim_{n\to \infty}\frac{n^{45}(1+\frac{2}{n})^{50}(1-\frac{3}{n})^{80}}{(2-\frac{1}{n})^{40}.(3-\frac{2}{n})^{45}}\)
\(=\frac{\lim_{n\to \infty}[n^{45}(1+\frac{2}{n})^{50}(1-\frac{3}{n})^{80}]}{\lim_{n\to \infty}[(2-\frac{1}{n})^{40}.(3-\frac{2}{n})^{45}]}\)
\(=\frac{\lim_{n\to \infty}n^{45}.1^{50}.1^{80}}{2^{40}.3^{45}}=\frac{\infty}{2^{40}.3^{45}}=\infty\)
PT
2 tháng 5 2017
a)lim\(\dfrac{4-3^n}{2.3^n+2}\)=lim\(\dfrac{4.\dfrac{1^n}{3^n}-\dfrac{3^n}{3^n}}{2.\dfrac{3^n}{3^n}+2\dfrac{1^n}{3^n}}=\)\(lim\dfrac{4.(\dfrac{1}{3})^n-1}{2.1+2.(\dfrac{1}{3})^n}=\dfrac{4.0-1}{\dfrac{2+2.0}{ }}=\dfrac{-1}{2}\)
b) lim\(\dfrac{3^{n+1}-2^n}{2-2.3^n}=lim\dfrac{3^n.3-2^n}{2-2.3^n}=lim\dfrac{3.\dfrac{3^n}{3^n}-\left(\dfrac{2}{3}\right)^n}{2.\left(\dfrac{1}{3}\right)^n-2.\dfrac{3^n}{3^n}}=\dfrac{3.1-0}{2.0-2.1}=\dfrac{-3}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 1 2019
a/
\(lim\dfrac{\sqrt{n^2-n}-n}{n}=lim\dfrac{-n}{n\left(\sqrt{n^2-n}+n\right)}=lim\dfrac{-\dfrac{1}{n}}{1\left(\sqrt{1-\dfrac{1}{n}}+1\right)}=\dfrac{0}{2}=0\)
b/
\(lim\dfrac{2^n-5^{n+2}}{5^n-4^{2-n}}=lim\dfrac{8^n-25.20^n}{20^n-4^2}=lim\dfrac{\left(\dfrac{8}{20}\right)^n-25}{1-\dfrac{16}{20^n}}=\dfrac{0-25}{1-0}=-25\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 4 2020
\(lim\left(u_n\right)=lim\left(\frac{n}{n^2+1}\right)=lim\left(\frac{\frac{1}{n}}{1+\frac{1}{n^2}}\right)=\frac{0}{1}=0\)
b/
\(-1\le cos\frac{\pi}{n}\le1\Rightarrow-\frac{n}{n^2+1}\le v_n\le\frac{n}{n^2+1}\)
Mà \(lim\left(-\frac{n}{n^2+1}\right)=lim\left(\frac{n}{n^2+1}\right)=0\)
\(\Rightarrow lim\left(v_n\right)=0\)
1 tháng 2 2018
a) lim= - 1/0 = - vô cùng
d) lim x(x^99-2)+1/ x(x^49-2)+1 =lim (x^99-2)/(x^49-2)=1
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 1 2019
\(lim\dfrac{\left(2-n\right)\left(3+2n^3\right)}{2n^2-1}=lim\dfrac{\left(\dfrac{2}{n}-1\right)\left(\dfrac{3}{n}+2n^2\right)}{2-\dfrac{1}{n^2}}=-\infty\)
\(\dfrac{lim\left(\sqrt{4n^2+1}-2n\right)n}{\sqrt[3]{4-n^3}+n}=lim\dfrac{n\left(\sqrt[3]{\left(4-n^3\right)^2}-n\sqrt[3]{4-n^3}+n^2\right)}{4.\left(\sqrt{4n^2+1}+2n\right)}\)
\(=lim\dfrac{\sqrt[3]{\left(n^3-4\right)^2}+n\sqrt[3]{n^3-4}+n^2}{4\left(\sqrt{4+\dfrac{1}{n^2}}+2\right)}=+\infty\)
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
15 tháng 5 2019
\(\lim\limits_{x\rightarrow-\infty}\frac{-x\sqrt{4x^2+3}}{2x-1}=\lim\limits_{x\rightarrow-\infty}\frac{x\sqrt{4+\frac{3}{x^2}}}{2-\frac{1}{x}}=-\infty\)
\(lim\frac{\sqrt{n}}{\sqrt{n+4}+\sqrt{n+3}}=lim\frac{1}{\sqrt{1+\frac{4}{n}}+\sqrt{1+\frac{3}{n}}}=\frac{1}{2}\)
\(lim\left(\frac{\left(n-2\right)^2-\left(3n^2+n-1\right)}{n-2+\sqrt{3n^2+n-1}}\right)=lim\frac{-2n^2-5n+5}{n-2+\sqrt{3n^2+n-1}}=lim\frac{-2n+5+\frac{5}{n}}{1-\frac{2}{n}+\sqrt{3+\frac{1}{n}-\frac{1}{n^2}}}=-\infty\)
\(\lim\limits_{x\rightarrow0}\frac{\left(x^3-2x+1\right)^{\frac{1}{3}}-1}{x^2+2x}=\lim\limits_{x\rightarrow0}\frac{\frac{1}{3}\left(3x-2\right)\left(x^3-2x+1\right)^{-\frac{2}{3}}}{2x+2}=-\frac{1}{3}\)
\(2^n=\left(1+1\right)^2=1+C_n^1+C_n^2+C_n^3+...+C_n^n>C_n^3\) (khi n đủ lớn)
\(\Rightarrow2^n>\dfrac{n\left(n-1\right)\left(n-2\right)}{6}\)
\(\Rightarrow\dfrac{n^2}{2^n}< \dfrac{6n^2}{n\left(n-1\right)\left(n-2\right)}=\dfrac{6n}{\left(n-1\right)\left(n-2\right)}\)
Đồng thời do \(\left\{{}\begin{matrix}n^2>0\\2^n>0\end{matrix}\right.\) \(\Rightarrow\dfrac{n^2}{2^n}>0\)
\(\Rightarrow0< \dfrac{n^2}{2^n}< \dfrac{6n}{\left(n-1\right)\left(n-2\right)}\)
Mà \(\lim\left(0\right)=\lim\left(\dfrac{6n}{\left(n-1\right)\left(n-2\right)}\right)=0\)
\(\Rightarrow\lim\left(\dfrac{n^2}{2^n}\right)=0\)