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

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

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

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

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

\(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{n\left(n+1\right)}=1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}\)

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

=> \(lim\left(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{n\left(n+1\right)}\right)=lim\left(2-\frac{1}{n+1}\right)=2\)( khi n tiến tới vô cùng )

Ta có: \(p\left(1\right)=1\)=>\(p\left(2\right)=p\left(1\right)=1\)

Với n > 2

\(p\left(n\right)=p\left(1\right)+2p\left(2\right)+...+\left(n-2\right)p\left(n-2\right)+\left(n-1\right)p\left(n-1\right)\)

=> \(p\left(n-1\right)=p\left(1\right)+2p\left(2\right)+...+\left(n-2\right)p\left(n-2\right)\)

=> \(p\left(n\right)-p\left(n-1\right)=\left(n-1\right)p\left(n-1\right)\)

=> \(p\left(n\right)=np\left(n-1\right)\)

Cứ thế tiếp tục: 

=> \(p\left(n\right)=np\left(n-1\right)=n.\left(n-1\right)p\left(n-2\right)=n\left(n-1\right).\left(n-2\right)...3.p\left(2\right)\)

\(=n\left(n-1\right).\left(n-2\right)...4.3\)