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

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

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

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

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

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

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

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

Ok cau nay em bi ngu :v No ko phai dang vo dinh ma nhan lien hop nhu dung roi :v Ma

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

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

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

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

Lạ nhỉ, tui chả biết dạng này dạng gì nữa :D

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

Cậu ơi :( Cậu chụp cái đề lên được ko, khó hịu thực sự :( 

\(C=\lim\limits\dfrac{\sqrt{n^2+1}}{n+1}?\)

\(C=\lim\limits\dfrac{\sqrt{\dfrac{n^2}{n^2}+\dfrac{1}{n^2}}}{\dfrac{n}{n}+\dfrac{1}{n}}=1\)