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a/ \(=lim\frac{1}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\infty}=0\)
b/ \(=lim\frac{6n+1}{\sqrt{n^2+5n+1}+\sqrt{n^2-n}}=\frac{6+\frac{1}{n}}{\sqrt{1+\frac{5}{n}+\frac{1}{n^2}}+\sqrt{1-\frac{1}{n}}}=\frac{6}{1+1}=3\)
c/ \(=lim\frac{6n-9}{\sqrt{3n^2+2n-1}+\sqrt{3n^2-4n+8}}=lim\frac{6-\frac{9}{n}}{\sqrt{3+\frac{2}{n}-\frac{1}{n^2}}+\sqrt{3-\frac{4}{n}+\frac{8}{n^2}}}=\frac{6}{\sqrt{3}+\sqrt{3}}=\sqrt{3}\)
d/ \(=lim\frac{\left(\frac{2}{6}\right)^n+1-4\left(\frac{4}{6}\right)^n}{\left(\frac{3}{6}\right)^n+6}=\frac{1}{6}\)
e/ \(=lim\frac{\left(\frac{3}{5}\right)^n-\left(\frac{4}{5}\right)^n+1}{\left(\frac{3}{5}\right)^n+\left(\frac{4}{5}\right)^n-1}=\frac{1}{-1}=-1\)
f/ Ta có công thức:
\(1+3+...+\left(2n+1\right)^2=\left(n+1\right)^2\)
\(\Rightarrow lim\frac{1+3+...+2n+1}{3n^2+4}=lim\frac{\left(n+1\right)^2}{3n^2+4}=lim\frac{\left(1+\frac{1}{n}\right)^2}{3+\frac{4}{n^2}}=\frac{1}{3}\)
g/ \(=lim\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}\right)=lim\left(1-\frac{1}{n+1}\right)=1-0=1\)
h/ Ta có: \(1^2+2^2+...+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)
\(\Rightarrow lim\frac{n\left(n+1\right)\left(2n+1\right)}{6n\left(n+1\right)\left(n+2\right)}=lim\frac{2n+1}{6n+12}=lim\frac{2+\frac{1}{n}}{6+\frac{12}{n}}=\frac{2}{6}=\frac{1}{3}\)
lim \(\frac{n\left(\sqrt[3]{2-n^3}+n\right)}{\sqrt{n^2+1}-n}\)
= lim \(\frac{n.2.\left(\sqrt{n^2+1}+n\right)}{\text{}\sqrt[3]{\left(2-n^3\right)^2}-n\sqrt[3]{2-n^3}+n^2}\)
= lim \(\frac{.2.\left(\sqrt{1+\frac{1}{n^2}}+1\right)}{\text{}\sqrt[3]{\left(\frac{2}{n^3}-1\right)^2}-\sqrt[3]{\frac{2}{n^3}-1}+1}\)
= \(\frac{2.\left(1+1\right)}{1+1+1}=\frac{4}{3}\)
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
\(a=lim\frac{n^2+n}{6n^3}=lim\frac{\frac{1}{n}+\frac{1}{n^3}}{6}=\frac{0}{6}=0\)
\(b=lim\frac{1+\frac{2}{n}}{1+\frac{1}{n}}+lim\frac{sinn}{2^n}=1+0=1\)
Giải thích: \(-1\le sin\left(n\right)\le1\) \(\forall n\Rightarrow\frac{-1}{2^n}\le\frac{sin\left(n\right)}{2^n}\le\frac{1}{2^n}\)
Mà \(lim\frac{-1}{2^n}=lim\frac{1}{2^n}=0\Rightarrow lim\frac{sin\left(n\right)}{2^n}=0\) theo nguyên tắc giới hạn kẹp
\(c=lim\frac{-3n-1}{\sqrt{n^2-3n}+\sqrt{n^2+1}}=lim\frac{-3-\frac{1}{n}}{\sqrt{1-\frac{3}{n}}+\sqrt{1+\frac{1}{n^2}}}=\frac{-3}{1+1}=-\frac{3}{2}\)
\(d=lim\frac{3n^2}{\sqrt[3]{\left(n^3+3n^2\right)^2}+n\sqrt[3]{n^3+3n^2}+n^2}=lim\frac{3}{\sqrt[3]{\left(1+\frac{3}{n}\right)^2}+\sqrt[3]{1+\frac{3}{n}}+1}=\frac{3}{1+1+1}=1\)
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?
\(=lim\frac{2.2^{5n}+3}{9.3^{5n}+1}=lim\frac{2.\left(\frac{2}{3}\right)^{5n}+3\left(\frac{1}{3}\right)^{5n}}{9+\left(\frac{1}{3}\right)^{5n}}=\frac{0}{9}=0\)
\(b=lim\frac{\left(-\frac{1}{3}\right)^n+4}{-1\left(-\frac{1}{3}\right)^n-2}=\frac{4}{-2}=-2\)
\(c=1+lim\frac{-n}{n^2+\sqrt{n^4+n}}=1+lim\frac{-\frac{1}{n}}{1+\sqrt{1+\frac{1}{n^3}}}=1+\frac{0}{2}=1\)
\(-2\le2cosn^2\le2\Rightarrow\frac{-2}{n^2+1}\le\frac{2cosn^2}{n^2+1}\le\frac{2}{n^2+1}\)
Mà \(lim\frac{-2}{n^2+1}=lim\frac{2}{n^2+1}=0\Rightarrow lim\frac{2cosn^2}{n^2+1}=0\)
\(d=lim\left[n\left(\sqrt{1-\frac{2}{n^2}}-1+1-\sqrt[3]{1+\frac{2}{n^2}}\right)\right]\)
\(=lim\left[n\left(\frac{-\frac{2}{n^2}}{\sqrt{1-\frac{2}{n^2}}+1}-\frac{\frac{2}{n^2}}{\sqrt[3]{\left(1+\frac{2}{n^2}\right)^2}+\sqrt[3]{1+\frac{2}{n^2}}+1}\right)\right]\)
\(=lim\left(\frac{-\frac{2}{n}}{\sqrt{1-\frac{2}{n^2}}+1}-\frac{\frac{2}{n}}{\sqrt[3]{\left(1+\frac{2}{n^2}\right)^2}+\sqrt[3]{1+\frac{2}{n^2}}+1}\right)=\frac{0}{2}-\frac{0}{1+1+1}=0\)
\(a=\lim\limits n\left(\sqrt[3]{\frac{1}{n}+1}+1\right)=+\infty.2=+\infty\)
\(b=\lim\limits\frac{n^2+2\sqrt{n}+3}{2n^2+n-\sqrt{n}}=\lim\limits\frac{1+\frac{2}{n\sqrt{n}}+\frac{3}{n^2}}{2+\frac{1}{n}-\frac{1}{n\sqrt{n}}}=\frac{1}{2}\)
\(c=\lim\limits\frac{2n\sqrt{n}+3}{n^2+n+1}=\frac{\frac{2}{\sqrt{n}}+\frac{3}{n^2}}{1+\frac{1}{n}+\frac{1}{n^2}}=\frac{0}{1}=0\)
\(d=\lim\limits\frac{2n^2+6n\sqrt{n}}{n^2+3n+2}=\lim\limits\frac{2+\frac{6}{\sqrt{n}}}{1+\frac{3}{n}+\frac{2}{n^2}}=\frac{2}{1}=2\)
\(\frac{1}{n\sqrt{n+1}+\left(n+1\right)\sqrt{n}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
\(\Rightarrow S_n=1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
\(\Rightarrow S_n=1-\frac{1}{\sqrt{n+1}}\)
\(lim\left(S_n\right)=lim\left(1-\frac{1}{\sqrt{n+1}}\right)=1-0=1\)
Lời giải:
Xét hạng tử tổng quát:
\(\frac{1}{(n+1)\sqrt{n}+n\sqrt{n+1}}=\frac{(n+1)-n}{\sqrt{n(n+1)}(\sqrt{n}+\sqrt{n+1)}}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n(n+1)}}\)
\(=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Cho $n=1,2,...$ thì:
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}=1-\frac{1}{\sqrt{2}}\)
\(\frac{1}{3\sqrt{2}+2\sqrt{3}}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\)
......
\(\frac{1}{(n+1)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
\(\Rightarrow U_n=1-\frac{1}{\sqrt{n+1}}\)
\(\Rightarrow \lim\limits U_n=\lim (1-\frac{1}{\sqrt{n+1}})=1\)