Tính các giới hạn sau:

\(\lim \frac{{\sqrt {4{n^2} + n + 1} }}{{8n + 3}}\);

\(\lim \frac{{\sqrt {4{n^2} + n + 1} }}{{8n + 3}}\)\( = \lim \frac{{\sqrt {{n^2}\left( {4 + \frac{1}{n} + \frac{1}{{{n^2}}}} \right)} }}{{n\left( {8 + \frac{3}{n}} \right)}}\)\( = \lim \frac{{\sqrt {4 + \frac{1}{n} + \frac{1}{{{n^2}}}} }}{{8 + \frac{3}{n}}}\)

\( = \frac{{\lim \left( {\sqrt {4 + \frac{1}{n} + \frac{1}{{{n^2}}}} } \right)}}{{\lim \left( {8 + \frac{3}{n}} \right)}} = \frac{{\sqrt {\lim \left( {4 + \frac{1}{n} + \frac{1}{{{n^2}}}} \right)} }}{{\lim \left( {8 + \frac{3}{n}} \right)}} = \frac{{\sqrt 4 }}{8} = \frac{2}{8} = \frac{1}{4}\).