Cai bai ben duoi bai nay y. Doc hieu chet lien. Ban nen xai go cong thuc de toi uu hon

\(C=\lim\limits\dfrac{n^3+1}{n\left(2n+1\right)^2}=\lim\limits\dfrac{n^3+1}{n\left(4n^2+4n+1\right)}=\lim\limits\dfrac{n^3+1}{4n^3+4n^2+n}=\lim\limits\dfrac{\dfrac{n^3}{n^3}+\dfrac{1}{n^3}}{\dfrac{4n^3}{n^3}+\dfrac{4n^2}{n^3}+\dfrac{n}{n^3}}=\dfrac{1}{4}\)

Xai cai nay go cong thuc di ban :v Doc ko hieu

\(E=\lim\limits\dfrac{\sqrt{n^3+2n}+1}{n+2}=\lim\limits\dfrac{\dfrac{\left(n^3+2n\right)^{\dfrac{1}{2}}}{n}+\dfrac{1}{n}}{\dfrac{n}{n}+\dfrac{2}{n}}=\dfrac{\dfrac{n^{\dfrac{3}{2}}}{n}}{\dfrac{n}{n}}=0\)

\(B=\lim\limits\left(\sqrt{2n^2+1}-n\right)?\)

\(B=\lim\limits\left[n\left(\sqrt{\dfrac{2n^2}{n^2}+\dfrac{1}{n^2}}-\dfrac{n}{n}\right)\right]=\lim\limits\left[n\left(\sqrt{2}-1\right)\right]=+\infty\)

\(M=\lim\limits\left(\sqrt[3]{1-n^2-8n^3}+2n\right)\)

\(=\lim\limits\dfrac{1-n^2-8n^3+8n^3}{\left(\sqrt[3]{1-n^2-8n^3}\right)^2-2n.\sqrt[3]{1-n^2-8n^3}+4n^2}\)

\(=\lim\limits\dfrac{1-n^2}{\left(1-n^2-8n^3\right)^{\dfrac{2}{3}}-2n.\left(1-n^2-8n^3\right)^{\dfrac{1}{3}}+4n^2}\)

\(=\lim\limits\dfrac{-\dfrac{n^2}{n^2}}{\dfrac{\left(-8n^3\right)^{\dfrac{2}{3}}}{n^2}-\dfrac{2n.\left(-8n^3\right)^{\dfrac{1}{3}}}{n^2}+\dfrac{4n^2}{n^2}}=\dfrac{-1}{4+4+4}=-\dfrac{1}{12}\)

\(F=\lim\limits\dfrac{\sqrt[4]{n^4-2n+1}+2n}{\sqrt[3]{3n^3+n}-n}=\lim\limits\dfrac{\sqrt[4]{\dfrac{n^4}{n^4}-\dfrac{2n}{n^4}+\dfrac{1}{n^4}}+\dfrac{2n}{n}}{\sqrt[3]{\dfrac{3n^3}{n^3}+\dfrac{n}{n^3}}-\dfrac{n}{n}}=\dfrac{1+2}{3-1}=\dfrac{3}{2}\)

Ta có: \(\sqrt{n^2 + 1}-\sqrt[3]{3n^3 + 2}\)

\(=n\sqrt{1 + \frac{1}{n^2}}-n\sqrt[3]{3 + \frac{2}{n^3}}\)

\(=n\left(\sqrt{1 + \frac{1}{n^2}}-\sqrt[3]{3 + \frac{2}{n^3}}\right)\)

Ta có: \(\sqrt[4]{2n^4 + n + 2}-n\)

\(=n\sqrt[4]{2 + \frac{1}{n^3} + \frac{2}{n^4}}-n\)

\(=n\left(\sqrt[4]{2 + \frac{1}{n^3} + \frac{2}{n^4}}-1\right)\)

Ta có: \(D = \lim_{n \to +\infty} \frac{\sqrt{n^2 + 1} - \sqrt[3]{3n^3 + 2}}{\sqrt[4]{2n^4 + n + 2} - n}\)

\(=\lim_{n\to+\infty}\frac{n \left( \sqrt{1 + \frac{1}{n^2}} - \sqrt[3]{3 + \frac{2}{n^3}} \right)}{n \left( \sqrt[4]{2 + \frac{1}{n^3} + \frac{2}{n^4}} - 1 \right)}\)

\(=\lim_{n\to+\infty}\frac{\sqrt{1 + \frac{1}{n^2}} - \sqrt[3]{3 + \frac{2}{n^3}}}{\sqrt[4]{2 + \frac{1}{n^3} + \frac{2}{n^4}} - 1}\)

\(=\frac{\sqrt{1 + 0} - \sqrt[3]{3 + 0}}{\sqrt[4]{2 + 0 + 0} - 1}=\frac{1 - \sqrt[3]{3}}{\sqrt[4]{2} - 1}\)

\(\lim\dfrac{n^4-3n+4}{an^3+2n^2+1}=\lim\dfrac{n-\dfrac{3}{n^2}+\dfrac{4}{n^3}}{a+\dfrac{2}{n}+\dfrac{1}{n^3}}=+\infty.\left(\dfrac{1}{a}\right)\)

Giới hạn đã cho bằng \(-\infty\) khi và chỉ khi \(\dfrac{1}{a}< 0\Leftrightarrow a< 0\)

\(\lim\limits\left(\sqrt{n^2+2n}-\sqrt{n^2-2n}\right)\)

\(=\lim\limits\dfrac{n^2+2n-n^2+2n}{\sqrt{n^2+2n}+\sqrt{n^2-2n}}\)

\(=\lim\limits\dfrac{4n}{\sqrt{n^2+2n}+\sqrt{n^2-2n}}\)

\(=\lim\limits\dfrac{4}{\sqrt{1+\dfrac{2}{n}}+\sqrt{1-\dfrac{2}{n}}}\)

\(=\dfrac{4}{1+1}=\dfrac{4}{2}=2\)