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a/ Bạn coi lại đề bài, 3n^2 +n^2 thì bằng 4n^2 luôn chứ ko ai cho đề bài như vậy cả
b/ \(\lim\limits\dfrac{\dfrac{n^3}{n^3}+\dfrac{3n}{n^3}+\dfrac{1}{n^3}}{-\dfrac{n^3}{n^3}+\dfrac{2n}{n^3}}=-1\)
c/ \(=\lim\limits\dfrac{-\dfrac{2n^3}{n^2}+\dfrac{3n}{n^2}+\dfrac{1}{n^2}}{-\dfrac{n^2}{n^2}+\dfrac{n}{n^2}}=\lim\limits\dfrac{-2n}{-1}=+\infty\)
d/ \(=\lim\limits\left[n\left(1+1\right)\right]=+\infty\)
e/ \(\lim\limits\left[2^n\left(\dfrac{2n}{2^n}-3+\dfrac{1}{2^n}\right)\right]=\lim\limits\left(-3.2^n\right)=-\infty\)
f/ \(=\lim\limits\dfrac{4n^2-n-4n^2}{\sqrt{4n^2-n}+2n}=\lim\limits\dfrac{-\dfrac{n}{n}}{\sqrt{\dfrac{4n^2}{n^2}-\dfrac{n}{n^2}}+\dfrac{2n}{n}}=-\dfrac{1}{2+2}=-\dfrac{1}{4}\)
g/ \(=\lim\limits\dfrac{n^2+3n-1-n^2}{\sqrt{n^2+3n-1}+n}+\lim\limits\dfrac{n^3-n^3+n}{\sqrt[3]{\left(n^3-n\right)^2}+n.\sqrt[3]{n^3-n}+n^2}\)
\(=\lim\limits\dfrac{\dfrac{3n}{n}-\dfrac{1}{n}}{\sqrt{\dfrac{n^2}{n^2}+\dfrac{3n}{n^2}-\dfrac{1}{n^2}}+\dfrac{n}{n}}+\lim\limits\dfrac{\dfrac{n}{n^2}}{\dfrac{\sqrt[3]{\left(n^3-n\right)^2}}{n^2}+\dfrac{n\sqrt[3]{n^3-n}}{n^2}+\dfrac{n^2}{n^2}}\)
\(=\dfrac{3}{2}+0=\dfrac{3}{2}\)
\(a,lim\dfrac{2n^2+1}{3n^3-3n+3}\)
\(=lim\dfrac{\dfrac{2}{n}+\dfrac{1}{n^3}}{3-\dfrac{3}{n^2}+\dfrac{3}{n^3}}=0\)
\(\lim\dfrac{-3n^3+1}{2n+5}=\lim\dfrac{-3n^2+\dfrac{1}{n}}{2+\dfrac{5}{n}}=\dfrac{-\infty}{2}=-\infty\)
\(\lim\dfrac{n^3-2n+1}{-3n-4}=\lim\dfrac{n^2-2+\dfrac{1}{n}}{-3-\dfrac{4}{n}}=\dfrac{+\infty}{-3}=-\infty\)
Chụp ảnh hoặc sử dụng gõ công thức nhé bạn. Để vầy khó hiểu lắm
\(a=\lim4^n\left(1-\left(\dfrac{3}{4}\right)^n\right)=+\infty.1=+\infty\)
\(b=\lim\left(4^n+2.2^n+1-4^n\right)=\lim2^n\left(2+\dfrac{1}{2^n}\right)=+\infty.2=+\infty\)
\(c=limn^3\left(\sqrt{\dfrac{2}{n}-\dfrac{3}{n^4}+\dfrac{11}{n^6}}-1\right)=+\infty.\left(-1\right)=-\infty\)
\(d=\lim n\left(\sqrt{2+\dfrac{1}{n^2}}-\sqrt{3-\dfrac{1}{n^2}}\right)=+\infty\left(\sqrt{2}-\sqrt{3}\right)=-\infty\)
\(e=\lim\dfrac{3n\sqrt{n}+1}{\sqrt{n^2+3n\sqrt{n}+1}+n}=\lim\dfrac{3\sqrt{n}+\dfrac{1}{n}}{\sqrt{1+\dfrac{3}{\sqrt{n}}+\dfrac{1}{n^2}}+1}=\dfrac{+\infty}{2}=+\infty\)
\(a=\lim n\left(\sqrt[3]{-1+\dfrac{2}{n}-\dfrac{5}{n^3}}\right)=+\infty.\left(-1\right)=-\infty\)
\(b=\lim\left(\sqrt{n+1}+\sqrt{n}\right)=+\infty\)
\(c=\lim n\left(\dfrac{1}{n^2+n}-1\right)=+\infty.\left(-1\right)=-\infty\)
\(d=\lim\left(\dfrac{2n^2-1-2n\left(n+1\right)}{n+1}\right)=\lim\left(\dfrac{-1-2n}{n+1}\right)=-2\)
\(e=\lim\dfrac{2n^2+n-3+\dfrac{1}{n}}{\dfrac{2}{n}-3}=\dfrac{+\infty}{-3}=-\infty\)
\(a=\lim\dfrac{5n\left(n+\sqrt{n^2-n-1}\right)}{n+1}=\lim\dfrac{5\left(n+\sqrt{n^2-n-1}\right)}{1+\dfrac{1}{n}}=\dfrac{+\infty}{1}=+\infty\)
\(b=\lim\dfrac{\sqrt{\dfrac{1}{n}+\sqrt{\dfrac{1}{n^3}+\dfrac{1}{n^4}}}}{1-\dfrac{1}{\sqrt{n}}}=\dfrac{0}{1}=0\)
\(c=\lim\dfrac{\sqrt{2n^2-1+\dfrac{7}{n^2}}}{3+\dfrac{5}{n}}=\dfrac{+\infty}{3}=+\infty\)
\(d=\lim\dfrac{\sqrt{3+\dfrac{2}{n}}-1}{3-\dfrac{2}{n}}=\dfrac{\sqrt{3}-1}{3}\)
\(a=\lim\left(\dfrac{2n^3\left(5n+1\right)+\left(2n^2+3\right)\left(1-5n^2\right)}{\left(2n^2+3\right)\left(5n+1\right)}\right)\)
\(=\lim\left(\dfrac{2n^3-13n^2+3}{\left(2n^2+3\right)\left(5n+1\right)}\right)=\lim\dfrac{2-\dfrac{13}{n}+\dfrac{3}{n^3}}{\left(2+\dfrac{3}{n^2}\right)\left(5+\dfrac{1}{n}\right)}=\dfrac{2}{2.5}=\dfrac{1}{5}\)
\(b=\lim\left(\dfrac{n-2}{\sqrt{n^2+n}+\sqrt{n^2+2}}\right)=\lim\dfrac{1-\dfrac{2}{n}}{\sqrt{1+\dfrac{1}{n}}+\sqrt{1+\dfrac{2}{n}}}=\dfrac{1}{2}\)
\(c=\lim\dfrac{\sqrt{1+\dfrac{3}{n^3}-\dfrac{2}{n^4}}}{2-\dfrac{2}{n}+\dfrac{3}{n^2}}=\dfrac{1}{2}\)
\(d=\lim\dfrac{\sqrt{1-\dfrac{4}{n}}-\sqrt{4+\dfrac{1}{n^2}}}{\sqrt{3+\dfrac{1}{n^2}}-1}=\dfrac{1-2}{\sqrt{3}-1}=-\dfrac{1+\sqrt{3}}{2}\)
\(=lim\frac{n\sqrt{1+\frac{1}{n}-\frac{1}{n^2}}-n\sqrt{4-\frac{2}{n^2}}}{n\left(1+\frac{3}{n}\right)}=\frac{\sqrt{1+0+0}-\sqrt{4-0}}{1+0}=-1\)
\(=lim\frac{3\left(\frac{3}{7}\right)^n-\frac{1}{4}.\left(\frac{2}{7}\right)^n-5.\left(\frac{1}{7}\right)^n}{3+6.\left(\frac{1}{7}\right)^n}=\frac{3.0-\frac{1}{4}.0-5.0}{3+6.0}=0\)
\(=lim\frac{2n-4}{3n+\sqrt{9n^2-2n+4}}=lim\frac{2-\frac{4}{n}}{3+\sqrt{9-\frac{2}{n}+\frac{4}{n^2}}}=\frac{2}{3+\sqrt{9}}=\frac{1}{3}\)