woa ai ni

\(lim\frac{\sqrt{n^6-n+1}+n^2}{3n^2\sqrt{n^2-1}}=lim\frac{\sqrt{n^6-n+1}+n^2}{3\sqrt{n^6-n^4}}\)

\(=lim\frac{\sqrt{1-\frac{1}{n^5}+\frac{1}{n^6}}+\frac{1}{n}}{3\sqrt{1-\frac{1}{n^2}}}\)(chia cả tử và mẫu cho n³

\(=\frac{\sqrt{1-0+0}+0}{3\sqrt{1-0}}=\frac{1}{3}\)

\(lim\left(\sqrt{2n^2+1}+\sqrt{2n^2-1}\right)\)

\(=lim\left|n\right|\left(\sqrt{2+\frac{1}{n^2}}+\sqrt{2-\frac{1}{n^2}}\right)\)

Do \(lim\left|n\right|=+\infty\)

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

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

\(lim\frac{1+2\cdot3^n-7^n}{5^n+2\cdot7^n}\)

\(=lim\frac{\frac{1}{7^n}+\frac{6^n}{7^n}-1}{\frac{5^n}{7^n}+\frac{14^n}{7^n}}\)

\(=lim\frac{0+\left(\frac{6}{7}\right)^n-1}{\left(\frac{5}{7}\right)^n+2}=\frac{-1}{2}\)

\(lim\frac{\sqrt{4n^2+1}+2n-1}{\sqrt{n^2+4n+1}+n}\)

= \(lim\frac{\sqrt{4+\frac{1}{n^2}}+2-\frac{1}{n}}{\sqrt{1+\frac{4}{n}+\frac{1}{n^2}}+1}\)

=\(\frac{2+2}{1+1}=2\)

lim \(\frac{-3n^2+5n+1}{2n^2-n+3}\)

= lim \(\frac{-3+\frac{5}{n}+\frac{1}{n^2}}{2-\frac{1}{n}+\frac{3}{n^2}}\)

= -3/2

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

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

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

= \(\frac{2.\left(1+1\right)}{1+1+1}=\frac{4}{3}\)