Tính các giới hạn sau:

a) 

\(\mathop {\lim }\limits_{n \to + \infty } \frac{{\sqrt {9{n^4} + 3{n^2} + 1} }}{{4{n^2} + 2}} = \mathop {\lim }\limits_{n \to + \infty } \frac{{\sqrt {\frac{{9{n^4} + 3{n^2} + 1}}{{{n^4}}}} }}{{\frac{{4{n^2} + 2}}{{{n^2}}}}}\)

\(\mathop {\lim }\limits_{n \to + \infty } \frac{{\sqrt {9 + \frac{3}{{{n^2}}} + \frac{1}{{{n^4}}}} }}{{4 + \frac{2}{{{n^2}}}}} = \frac{3}{4}\)\(\)

b) 

\[\mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - 1}}{{{x^2} - 5x + 4}} = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x - 1} \right)\left( {x + 1} \right)}}{{(x - 1)(x - 4)}}\]

\[ = \mathop {\lim }\limits_{x \to 1} \frac{{x + 1}}{{x - 4}} = \frac{{ - 2}}{3}\]