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\(lim\dfrac{\left(n+2\right)^{50}\left(n-3\right)^{80}}{\left(2n-1\right)^{40}\left(3n-2\right)^{45}}=lim\dfrac{\left(1+\dfrac{2}{n^{50}}\right)\left(1-\dfrac{3}{n^{35}}\right)\left(n-3\right)^{45}}{\left(2-\dfrac{1}{n^{50}}\right)\left(3-\dfrac{2}{n^{45}}\right)}=+\infty\)
\(lim\dfrac{4^n}{2.3^n+4^n}=lim\dfrac{1}{2.\left(\dfrac{3}{4}\right)^n+1}=\dfrac{1}{0+1}=1\)
\(lim\dfrac{3^n-2.5^n}{7+3.5^n}=lim\dfrac{\left(\dfrac{3}{5}\right)^n-2}{\dfrac{7}{5^n}+3}=\dfrac{0-2}{0+3}=\dfrac{-2}{3}\)
\(lim\dfrac{4^n-5^n}{2^{2n}+3.5^{2n}}=lim\dfrac{\left(\dfrac{4}{25}\right)^n-\left(\dfrac{1}{5}\right)^n}{\left(\dfrac{2}{5}\right)^{2n}+3}=\dfrac{0-0}{0+3}=0\)
\(lim\dfrac{\left(-3\right)^n+5^n}{2.\left(-4\right)^n+5^n}=lim\dfrac{\left(\dfrac{-3}{5}\right)^n+1}{2.\left(-\dfrac{4}{5}\right)^n+1}=\dfrac{0+1}{0+1}=1\)
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:
\(\lim_{n\to \infty}\frac{(n+2)^{50}.(n-3)^{80}}{(2n-1)^{40}.(3n-2)^{45}}=\lim_{n\to \infty}\frac{n^{130}(\frac{n+2}{n})^{50}.(\frac{n-3}{n})^{80}}{n^{85}(\frac{2n-1}{n})^{40}.(\frac{3n-2}{n})^{45}}\)
\(=\lim_{n\to \infty}\frac{n^{45}(1+\frac{2}{n})^{50}(1-\frac{3}{n})^{80}}{(2-\frac{1}{n})^{40}.(3-\frac{2}{n})^{45}}\)
\(=\frac{\lim_{n\to \infty}[n^{45}(1+\frac{2}{n})^{50}(1-\frac{3}{n})^{80}]}{\lim_{n\to \infty}[(2-\frac{1}{n})^{40}.(3-\frac{2}{n})^{45}]}\)
\(=\frac{\lim_{n\to \infty}n^{45}.1^{50}.1^{80}}{2^{40}.3^{45}}=\frac{\infty}{2^{40}.3^{45}}=\infty\)
\(a=lim\dfrac{\left(\dfrac{2}{6}\right)^n+1-\dfrac{1}{4}\left(\dfrac{4}{6}\right)^n}{\left(\dfrac{3}{6}\right)^n+6}=\dfrac{1}{6}\)
\(b=\lim\dfrac{\left(n+1\right)^2}{3n^2+4}=\lim\dfrac{n^2+2n+1}{3n^2+4}=\lim\dfrac{1+\dfrac{2}{n}+\dfrac{1}{n^2}}{3+\dfrac{4}{n^2}}=\dfrac{1}{3}\)
\(c=\lim\dfrac{n\left(n+1\right)}{2\left(n^2-3\right)}=\lim\dfrac{n^2+n}{2n^2-6}=\lim\dfrac{1+\dfrac{1}{n}}{2-\dfrac{6}{n^2}}=\dfrac{1}{2}\)
\(d=\lim\left[1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\right]=\lim\left[1-\dfrac{1}{n+1}\right]=1\)
\(e=\lim\dfrac{1}{2}\left[1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right]\)
\(=\lim\dfrac{1}{2}\left[1-\dfrac{1}{2n+1}\right]=\dfrac{1}{2}\)
\(\lim\dfrac{\left(2n-1\right)\left(3n^2+2\right)^3}{-2n^5+4n^3-1}=\lim\dfrac{\left(\dfrac{2n-1}{n}\right)\left(\dfrac{3n^2+2}{n^2}\right)^3}{\dfrac{-2n^5+4n^3-1}{n^7}}\)
\(=\lim\dfrac{\left(2-\dfrac{1}{n}\right)\left(3+\dfrac{2}{n^2}\right)^3}{-\dfrac{2}{n^2}+\dfrac{4}{n^4}-\dfrac{1}{n^7}}=-\infty\)
\(\lim3^n\left(6.\left(\dfrac{2}{3}\right)^n-5+\dfrac{7n}{3^n}\right)=+\infty.\left(-5\right)=-\infty\)
\(lim\dfrac{\left(2-n\right)\left(3+2n^3\right)}{2n^2-1}=lim\dfrac{\left(\dfrac{2}{n}-1\right)\left(\dfrac{3}{n}+2n^2\right)}{2-\dfrac{1}{n^2}}=-\infty\)
\(\dfrac{lim\left(\sqrt{4n^2+1}-2n\right)n}{\sqrt[3]{4-n^3}+n}=lim\dfrac{n\left(\sqrt[3]{\left(4-n^3\right)^2}-n\sqrt[3]{4-n^3}+n^2\right)}{4.\left(\sqrt{4n^2+1}+2n\right)}\)
\(=lim\dfrac{\sqrt[3]{\left(n^3-4\right)^2}+n\sqrt[3]{n^3-4}+n^2}{4\left(\sqrt{4+\dfrac{1}{n^2}}+2\right)}=+\infty\)
1. Bạn ghi lại đề, mẫu số ko rõ
2. \(=lim\left[-8n^6\left(1-\frac{4}{n^2}\right)^3\right]=-\infty.1=-\infty\)
3. Dãy số là CSC với \(\left\{{}\begin{matrix}u_1=-1\\d=3\end{matrix}\right.\) \(\Rightarrow u_n=-1+\left(n-1\right)3=3n-4\)
\(\Rightarrow lim\frac{3n-4}{5n+2020}=lim\frac{3-\frac{4}{n}}{5+\frac{2020}{n}}=\frac{3}{5}\)
4.
\(u_{n+1}=\frac{1}{2}u_n+\frac{3}{2}\Rightarrow u_{n+1}-3=\frac{1}{2}\left(u_n-3\right)\)
Đặt \(v_n=u_n-3\Rightarrow\left\{{}\begin{matrix}v_1=-2\\v_{n+1}=\frac{1}{2}v_n\end{matrix}\right.\)
\(\Rightarrow v_n\) là CSN với công bội \(\frac{1}{2}\Rightarrow v_n=-2.\frac{1}{2^{n-1}}\Rightarrow u_n=v_n+3=-\frac{1}{2^{n-2}}+3\)
\(\Rightarrow lim\left(u_n\right)=lim\left[-\frac{1}{2^{n-2}}+3\right]=3\)
5.
\(u_{n+1}=u_n+\frac{1}{2^n}\Rightarrow u_{n+1}+\frac{2}{2^{n+1}}=u_n+\frac{2}{2^n}\)
Đặt \(v_n=u_n+\frac{2}{2^n}\Rightarrow\left\{{}\begin{matrix}v_1=3\\v_{n+1}=v_n\end{matrix}\right.\)
\(\Rightarrow v_{n+1}=v_n=...=v_1=3\Rightarrow u_n=3-\frac{2}{2^n}\)
\(\Rightarrow u_{n-2}=3-\frac{2}{2^{n-2}}\Rightarrow lim\left(u_{n-2}\right)=lim\left(3-\frac{2}{2^{n-2}}\right)=3\)
Tính \(u_{n-2}\) hay \(u_n-2\) nhỉ? Ko dịch nổi nên đoán đại
\(a=\lim n\left(\sqrt[3]{-1+\dfrac{2}{n}-\dfrac{5}{n^3}}\right)=+\infty.\left(-1\right)=-\infty\)
\(b=\lim\left(\sqrt{n+1}+\sqrt{n}\right)=+\infty\)
\(c=\lim n\left(\dfrac{1}{n^2+n}-1\right)=+\infty.\left(-1\right)=-\infty\)
\(d=\lim\left(\dfrac{2n^2-1-2n\left(n+1\right)}{n+1}\right)=\lim\left(\dfrac{-1-2n}{n+1}\right)=-2\)
\(e=\lim\dfrac{2n^2+n-3+\dfrac{1}{n}}{\dfrac{2}{n}-3}=\dfrac{+\infty}{-3}=-\infty\)
Bạn muốn tìm giới hạn nhưng lại không chỉ rõ $n$ chạy đến đâu?
Điển hình như câu 1:
$n\to 0$ thì giới hạn là $3$
$n\to \pm \infty$ thì giới hạn là $\pm \infty$
Bạn phải ghi rõ đề ra chứ?
\(\lim\left(1+\dfrac{-1}{2^n}\right)=1+0=1\Rightarrow a=1\)
\(\lim\left(\dfrac{n^5}{n^4-2n^3+1}-n\right)=\lim\left(\dfrac{n^5-n\left(n^4-2n^3+1\right)}{n^4-2n^3+1}\right)\)
\(=\lim\left(\dfrac{2n^4-n}{n^4-2n^3+1}\right)=\lim\left(\dfrac{2-\dfrac{1}{n^3}}{1-\dfrac{2}{n}+\dfrac{1}{n^4}}\right)=2\)