Tính các giới hạn sau:

a) \[\lim \frac{{\sqrt[{}]{{4{n^2} + 3n + 1}}}}{{2n - 1}}\]\[ = \lim \frac{{\sqrt[{}]{{{n^2}\left( {4 + \frac{3}{n} + \frac{1}{{{n^2}}}} \right)}}}}{{n\left( {2 - \frac{1}{n}} \right)}}\]\[ = \lim \frac{{n.\sqrt[{}]{{4 + \frac{3}{n} + \frac{1}{{{n^2}}}}}}}{{n\left( {2 - \frac{1}{n}} \right)}}\]\[ = \lim \frac{{\sqrt[{}]{{4 + \frac{3}{n} + \frac{1}{{{n^2}}}}}}}{{2 - \frac{1}{n}}} = 1\].

b)\[\mathop {\lim }\limits_{x \to 1} \left[ {\frac{{{x^3} - 3{x^2} + 3x - 1}}{{{{(x - 1)}^2}}} + \frac{{x + 2 - \sqrt[{}]{{6x + 3}}}}{{{{(x - 1)}^2}}}} \right] = \mathop {\lim }\limits_{x \to 1} \left[ {\frac{{{{(x - 1)}^3}}}{{{{(x - 1)}^2}}} + \frac{{{x^2} - 2x + 1}}{{{{(x - 1)}^2}(x + 2 + \sqrt[{}]{{6x + 3}})}}} \right]\]

\[ = \mathop {\lim }\limits_{x \to 1} \left[ {x - 1 + \frac{1}{{x + 2 + \sqrt[{}]{{6x + 3}}}}} \right] = \frac{1}{6}\].