Tính các giới hạn sau: lim ((căn bậc hai (9n^2 + 2n + 1)) / (n - 5))
\(\lim \frac{{\sqrt {9{n^2} + 2n + 1} }}{{n - 5}}\);
\(\lim \frac{{\sqrt {9{n^2} + 2n + 1} }}{{n - 5}}\)\( = \lim \frac{{\sqrt {{n^2}\left( {9 + \frac{2}{n} + \frac{1}{{{n^2}}}} \right)} }}{{n - 5}}\)\( = \lim \frac{{n\sqrt {9 + \frac{2}{n} + \frac{1}{{{n^2}}}} }}{{n\left( {1 - \frac{5}{n}} \right)}}\)
\( = \lim \frac{{\sqrt {9 + \frac{2}{n} + \frac{1}{{{n^2}}}} }}{{1 - \frac{5}{n}}} = \frac{{\lim \sqrt {9 + \frac{2}{n} + \frac{1}{{{n^2}}}} }}{{\lim \left( {1 - \frac{5}{n}} \right)}} = \frac{{\sqrt 9 }}{1} = 3\).