1) tính giới hạn \(\lim\limits_{n\rightarrow\infty}\sqrt{n^2-1}+3n\)
2) tính giới hạn I = \(\lim\limits_{n\rightarrow\infty}\left(\sqrt{4n^2+5}+n\right)\)
1:
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2-1-9n^2}{\sqrt{n^2-1}-3n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{-8n^2-1}{\sqrt{n^2-1}-3n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2\left(-8-\dfrac{1}{n^2}\right)}{n\left(\sqrt{1-\dfrac{1}{n^2}}-3\right)}=\lim\limits_{n\rightarrow\infty}-\dfrac{8}{1-3}\cdot n=\lim\limits_{n\rightarrow\infty}4n=+\infty\)
2:
\(\lim\limits_{n\rightarrow\infty}\sqrt{4n^2+5}+n\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{4n^2+5-n^2}{\sqrt{4n^2+5}-n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{3n^2+5}{\sqrt{4n^2+5}-n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2\left(3+\dfrac{5}{n^2}\right)}{n\left(\sqrt{4+\dfrac{5}{n^2}}-1\right)}\)
\(=\lim\limits_{n\rightarrow\infty}n\cdot\left(\dfrac{3}{\sqrt{4}-1}\right)=+\infty\)
1) Tính giới hạn \(\lim\limits_{n\rightarrow\infty}\left(\sqrt[3]{n^3+3n^2+1}-n\right)\)
2) Tính giới hạn \(\lim\limits_{n\rightarrow\infty}\left(\sqrt{4n^2+1}-\sqrt[3]{8n^3+n}\right)\)
1:
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^3+3n^2+1-n^3}{\sqrt[3]{n^3+3n^2+1}+n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{3n^2+1}{\sqrt[3]{n^3+3n^2+1}+n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2\left(3+\dfrac{1}{n^2}\right)}{n\left(\sqrt[3]{1+\dfrac{3}{n}+\dfrac{1}{n^3}}+1\right)}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n\cdot\left(3+\dfrac{1}{n^2}\right)}{\sqrt[3]{1+\dfrac{3}{n}+\dfrac{1}{n^3}}+1}\)
\(=\lim\limits_{n\rightarrow\infty}n\cdot\lim\limits_{n\rightarrow\infty}\dfrac{3+\dfrac{1}{n^2}}{\sqrt[3]{1+\dfrac{3}{n}+\dfrac{1}{n^3}}+1}\)
\(=+\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{n\rightarrow\infty}n=+\infty\\\lim\limits_{n\rightarrow\infty}\dfrac{3+\dfrac{1}{n^2}}{\sqrt[3]{1+\dfrac{3}{n}+\dfrac{1}{n^3}}+1}=\dfrac{3}{2}>0\end{matrix}\right.\)
2:
\(=\lim\limits_{n\rightarrow\infty}\left(\sqrt{4n^2+1}-2n+2n-\sqrt[3]{8n^3+n}\right)\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{4n^2+1-4n^2}{\sqrt{4n^2+1}+2n}+\lim\limits_{n\rightarrow\infty}\dfrac{8n^3-8n^3-n}{4n^2+2n\cdot\sqrt[3]{8n^3+n}+\left(\sqrt[3]{8n^3+n}\right)^2}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{1}{\sqrt{4n^2+1}+2n}+\lim\limits_{n\rightarrow\infty}\dfrac{-n}{4n^2+2n\cdot n\cdot\sqrt[3]{8+\dfrac{1}{n^3}}+\left(n\cdot\sqrt[3]{8+\dfrac{1}{n^3}}\right)^2}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{-n}{4n^2+2n^2\cdot\sqrt[3]{8+\dfrac{1}{n^3}}+n^2\cdot\sqrt[3]{\left(8+\dfrac{1}{n^3}\right)^2}}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{-1}{4n+2n\cdot\sqrt[3]{8+\dfrac{1}{n^3}}+n\cdot\sqrt[3]{\left(8+\dfrac{1}{n^3}\right)^2}}\)
\(=0\)
1) Tính giới hạn \(\lim\limits_{n\rightarrow\infty}\left(\dfrac{3^n-4^{n+1}}{3^{n+2}+4^n}\right)\)
2) Tính giới hạn \(\lim\limits_{n\rightarrow\infty}\left(\dfrac{3^n-4.2^{n+1}-3}{3.2^n+4^n}\right)\)
3) Tính giới hạn \(\lim\limits_{n\rightarrow\infty}\left(\dfrac{2-5^{n-2}}{3^n+2.5^n}\right)\)
3:
\(\lim\limits_{n\rightarrow\infty}\dfrac{2-5^{n-2}}{3^n+2\cdot5^n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{\dfrac{2}{5^n}-\dfrac{5^{n-2}}{5^n}}{\dfrac{3^n}{5^n}+2\cdot\dfrac{5^n}{5^n}}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{\dfrac{2}{5^n}-\dfrac{1}{25}}{\left(\dfrac{3}{5}\right)^n+2\cdot1}\)
\(=-\dfrac{1}{25}:2=-\dfrac{1}{50}\)
1:
\(=\lim\limits_{n\rightarrow\infty}\dfrac{3^n-4^n\cdot4}{3^n\cdot9+4^n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{\dfrac{3^n}{4^n}-4}{3^n\cdot\dfrac{9}{4^n}+1}\)
\(=-\dfrac{4}{1}=-4\)
1) Tính giới hạn \(\lim\limits_{n\rightarrow\infty}\left(\dfrac{3^n-4^{n+1}}{3^{n+2}+4^n}\right)\)
2) Tính giới hạn \(\lim\limits_{n\rightarrow\infty}\left(\dfrac{3^n+1}{2^n-1}\right)\)
2:
\(\lim\limits_{n\rightarrow\infty}\dfrac{3^n+1}{2^n-1}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{\dfrac{3^n}{3^n}+\dfrac{1}{3^n}}{\dfrac{2^n}{3^n}-\dfrac{1}{3^n}}=\lim\limits_{n\rightarrow\infty}\dfrac{1+\dfrac{1}{3^n}}{\left(\dfrac{2}{3}\right)^n-\dfrac{1}{3^n}}=1\)
1) tính giới hạn \(\lim\limits_{n\rightarrow\infty}\left(\sqrt{n^2+6n}-n\right)\)
2) tính giới hạn \(\lim\limits_{n\rightarrow\infty}\left(\sqrt{n+1}-\sqrt{n-1}\right)\)
Lời giải:
1.
\(\lim\limits_{n\to \infty}(\sqrt{n^2+6n}-n)=\lim\limits_{n\to \infty}\frac{6n}{\sqrt{n^2+6n}+n}=\lim\limits_{n\to \infty}\frac{6}{\sqrt{1+\frac{6}{n}}+1}=\frac{6}{1+1}=3\)
2.
\(\lim\limits_{n\to \infty}(\sqrt{n+1}-\sqrt{n-1})=\lim\limits_{n\to \infty}\frac{(n+1)-(n-1)}{\sqrt{n+1}+\sqrt{n-1}}=\lim\limits_{n\to \infty}\frac{2}{\sqrt{n+1}+\sqrt{n-1}}=0\) do $\sqrt{n+1}+\sqrt{n-1}\to \infty$ khi $n\to \infty$
1) Tính giới hạn \(\lim\limits_{n\rightarrow\infty}\left(\dfrac{7^n+2^2}{3.7^n+4^n}\right)\)
2) Tính giới hạn \(\lim\limits_{n\rightarrow\infty}\left(\dfrac{1-4^n}{1+4^n}\right)\)
1:
\(\lim\limits_{n\rightarrow\infty}\dfrac{7^n+4}{3\cdot7^n+4^n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{1+\dfrac{4}{7^n}}{3+\left(\dfrac{4}{7}\right)^n}=\dfrac{1}{3}\)
2: \(\lim\limits_{n\rightarrow\infty}\dfrac{1-4^n}{1+4^n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{\dfrac{1}{4^n}-1}{\dfrac{1}{4^n}+1}=-\dfrac{1}{1}=-1\)
1) Tính giới hạn \(K=\lim\limits_{n\rightarrow\infty}\left(\dfrac{3.2^n-3^n}{2^{n+1}+3^{n+1}}\right)\)
2) Tính giới hạn \(\lim\limits_{n\rightarrow\infty}\left(\dfrac{3^n-4^{n+1}}{3^{n+2}+4^n}\right)\)
1:
\(K=\lim\limits_{n\rightarrow\infty}\dfrac{3\cdot2^n-3^n}{2^{n+1}+3^{n+1}}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{3\cdot2^n-3^n}{2^n\cdot2+3^n\cdot3}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{3\cdot\dfrac{2^n}{3^n}-1}{\left(\dfrac{2}{3}\right)^n\cdot2+3}\)
\(=-\dfrac{1}{3}\)
2:
\(\lim\limits_{n\rightarrow\infty}\dfrac{3^n-4^{n+1}}{3^{n+2}+4^n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{3^n-4^n\cdot4}{3^n\cdot9+4^n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{\left(\dfrac{3}{4}\right)^n-4}{\left(\dfrac{3}{4}\right)^n\cdot9+1}=-\dfrac{4}{1}=-4\)
1) Tính \(I=\lim\limits_{n\rightarrow\infty}\left(\sqrt{n^2+2}-\sqrt{n^2-1}\right)\)
2) Tính giới hạn \(\lim\limits_{n\rightarrow\infty}\left(\sqrt{n^2+2n+2}+n\right)\)
1: \(I=\lim\limits_{n\rightarrow\infty}\dfrac{n^2+2-n^2+1}{\sqrt{n^2+2}+\sqrt{n^2-1}}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{3}{\sqrt{n^2+2}+\sqrt{n^2-1}}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{3}{n\left(\sqrt{1+\dfrac{2}{n^2}}+\sqrt{1-\dfrac{1}{n^2}}\right)}\)
=0
2: \(\lim\limits_{n\rightarrow\infty}\sqrt{n^2+2n+2}+n\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2+2n+2-n^2}{\sqrt{n^2+2n+2}-n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{2n+2}{\sqrt{n^2+2n+2}-n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{2+\dfrac{1}{n}}{\sqrt{1+\dfrac{2}{n}+\dfrac{2}{n^2}}-1}\)
\(=+\infty\)
Tính các giới hạn
a) \(\lim\limits_{x\rightarrow\infty}\dfrac{a_0x^m+a_1x^{m-1}+a_2x^{m-2}+...+a_m}{b_0x^n+b_1x^{n-1}+b_2x^{n-2}+...+b_n}\)
b) \(\lim\limits_{x\rightarrow+\infty}\dfrac{\left(x-\sqrt{x^2-1}\right)^n+\left(x+\sqrt{x^2-1}\right)^n}{x^n}\)
