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\(a=lim\dfrac{\left(\dfrac{2}{6}\right)^n+1-\dfrac{1}{4}\left(\dfrac{4}{6}\right)^n}{\left(\dfrac{3}{6}\right)^n+6}=\dfrac{1}{6}\)
\(b=\lim\dfrac{\left(n+1\right)^2}{3n^2+4}=\lim\dfrac{n^2+2n+1}{3n^2+4}=\lim\dfrac{1+\dfrac{2}{n}+\dfrac{1}{n^2}}{3+\dfrac{4}{n^2}}=\dfrac{1}{3}\)
\(c=\lim\dfrac{n\left(n+1\right)}{2\left(n^2-3\right)}=\lim\dfrac{n^2+n}{2n^2-6}=\lim\dfrac{1+\dfrac{1}{n}}{2-\dfrac{6}{n^2}}=\dfrac{1}{2}\)
\(d=\lim\left[1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\right]=\lim\left[1-\dfrac{1}{n+1}\right]=1\)
\(e=\lim\dfrac{1}{2}\left[1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right]\)
\(=\lim\dfrac{1}{2}\left[1-\dfrac{1}{2n+1}\right]=\dfrac{1}{2}\)
\(a=\lim\left(\dfrac{2n^3\left(5n+1\right)+\left(2n^2+3\right)\left(1-5n^2\right)}{\left(2n^2+3\right)\left(5n+1\right)}\right)\)
\(=\lim\left(\dfrac{2n^3-13n^2+3}{\left(2n^2+3\right)\left(5n+1\right)}\right)=\lim\dfrac{2-\dfrac{13}{n}+\dfrac{3}{n^3}}{\left(2+\dfrac{3}{n^2}\right)\left(5+\dfrac{1}{n}\right)}=\dfrac{2}{2.5}=\dfrac{1}{5}\)
\(b=\lim\left(\dfrac{n-2}{\sqrt{n^2+n}+\sqrt{n^2+2}}\right)=\lim\dfrac{1-\dfrac{2}{n}}{\sqrt{1+\dfrac{1}{n}}+\sqrt{1+\dfrac{2}{n}}}=\dfrac{1}{2}\)
\(c=\lim\dfrac{\sqrt{1+\dfrac{3}{n^3}-\dfrac{2}{n^4}}}{2-\dfrac{2}{n}+\dfrac{3}{n^2}}=\dfrac{1}{2}\)
\(d=\lim\dfrac{\sqrt{1-\dfrac{4}{n}}-\sqrt{4+\dfrac{1}{n^2}}}{\sqrt{3+\dfrac{1}{n^2}}-1}=\dfrac{1-2}{\sqrt{3}-1}=-\dfrac{1+\sqrt{3}}{2}\)
\(\lim\limits\left(2-3n\right)^4\left(n+1\right)^3=\lim n^7\left(3-\dfrac{2}{n}\right)^4\left(1+\dfrac{1}{n}\right)^3=+\infty\)
\(\lim\left(\sqrt[3]{n+4}-\sqrt[3]{n+1}\right)=\lim\dfrac{3}{\sqrt[3]{\left(n+4\right)^2}+\sqrt[3]{\left(n+4\right)\left(n+1\right)}+\sqrt[3]{\left(n+1\right)^2}}=0\)
\(\lim\left(\sqrt[3]{8n^3+3n^2+4}-2n+6\right)=\lim\dfrac{8n^3+3n^2+4-\left(2n-6\right)^3}{\sqrt[3]{\left(8n^3+3n^2+4\right)^2}+\left(2n-6\right)\sqrt[3]{8n^3+3n^2+4}+\left(2n-6\right)^2}\)
\(=\lim\dfrac{75n^2-216n+220}{\sqrt[3]{\left(8n^3+3n^2+4\right)^2}+\left(2n-6\right)\sqrt[3]{8n^3+3n^2+4}+\left(2n-6\right)^2}\)
\(=\lim\dfrac{75-\dfrac{216}{n}+\dfrac{220}{n^2}}{\sqrt[3]{\left(8+\dfrac{3}{n}+\dfrac{4}{n^3}\right)^2}+\left(2-\dfrac{6}{n}\right)\sqrt[3]{8+\dfrac{3}{n}+\dfrac{4}{n^3}}+\left(2-\dfrac{6}{n}\right)^2}\)
\(=\dfrac{75}{\sqrt[3]{8^2}+2.\sqrt[3]{8}+2^2}=...\)
d.
\(\lim\left(\sqrt[3]{8n^3+3n^2-2}+\sqrt[3]{5n^2-8n^3}\right)\)
\(=\lim\left(\sqrt[3]{8n^3+3n^2-2}-\sqrt[3]{8n^3-5n^2}\right)\)
\(=\lim\dfrac{8n^3+3n^2-2-\left(8n^3-5n^2\right)}{\sqrt[3]{\left(8n^3+3n^2-2\right)^2}+\sqrt[3]{\left(8n^3+3n^2-2\right)\left(8n^3-5n^2\right)}+\sqrt[3]{8n^3-5n^2}}\)
\(=\lim\dfrac{8n^2-2}{\sqrt[3]{\left(8n^3+3n^2-2\right)^2}+\sqrt[3]{\left(8n^3+3n^2-2\right)\left(8n^3-5n^2\right)}+\sqrt[3]{8n^3-5n^2}}\)
\(=lim\dfrac{8-\dfrac{2}{n^2}}{\sqrt[3]{\left(8+\dfrac{3}{n}-\dfrac{2}{n^3}\right)^2}+\sqrt[3]{\left(8+\dfrac{3}{n}-\dfrac{2}{n^3}\right)\left(8-\dfrac{5}{n}\right)}+\sqrt[3]{\left(8-\dfrac{5}{n}\right)^2}}\)
\(=\dfrac{8}{\sqrt[3]{8^2}+\sqrt[3]{8.8}+\sqrt[3]{8^2}}=...\)
\(a=\lim4^n\left(1-\left(\dfrac{3}{4}\right)^n\right)=+\infty.1=+\infty\)
\(b=\lim\left(4^n+2.2^n+1-4^n\right)=\lim2^n\left(2+\dfrac{1}{2^n}\right)=+\infty.2=+\infty\)
\(c=limn^3\left(\sqrt{\dfrac{2}{n}-\dfrac{3}{n^4}+\dfrac{11}{n^6}}-1\right)=+\infty.\left(-1\right)=-\infty\)
\(d=\lim n\left(\sqrt{2+\dfrac{1}{n^2}}-\sqrt{3-\dfrac{1}{n^2}}\right)=+\infty\left(\sqrt{2}-\sqrt{3}\right)=-\infty\)
\(e=\lim\dfrac{3n\sqrt{n}+1}{\sqrt{n^2+3n\sqrt{n}+1}+n}=\lim\dfrac{3\sqrt{n}+\dfrac{1}{n}}{\sqrt{1+\dfrac{3}{\sqrt{n}}+\dfrac{1}{n^2}}+1}=\dfrac{+\infty}{2}=+\infty\)
Đáp án là A
l i m 1 1 . 3 + 1 2 . 4 + . . . + 1 n n + 2
Ta có :
l i m 1 1 . 3 + 1 2 . 4 + . . . + 1 n n + 2 = l i m 2 1 . 3 + 2 2 . 4 + . . . + 2 n n + 2
l i m 1 2 1 - 1 3 + 1 2 - 1 4 + 1 3 - 1 5 . . . + 1 n - 1 n + 2 = l i m 1 2 1 + 1 2 - 1 n + 2 = 3 4