Làm mấy câu gần đây thôi, rảnh thì làm thêm mấy câu xa xa chút

Mà t nhắn ko rep à

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

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

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

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

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

Cai nay dich ko noi :v

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

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

Lag :) Chu tus sua cai duoi mau la \(\dfrac{1}{n^2}\) nhe :v

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

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

\(\frac{4}{5}m^2=0,8m^2=80dm^2\)