Tính các giới hạn sau:

a)    \(\mathop {\lim }\limits_{x \to + \infty } \frac{{4{x^2} - 2x + 1}}{{ - 2 + x + 2{x^2}}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{4 - \frac{2}{x} + \frac{1}{{{x^2}}}}}{{ - \frac{2}{{{x^2}}} + \frac{1}{x} + 2}} = \frac{4}{2} = 2.\)

b) .

               \[\begin{array}{l}\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {x + 3} - 2}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{x + 3 - 4}}{{(x - 1)\left( {\sqrt {x + 3} + 2} \right)}}\\ = \mathop {\lim }\limits_{x \to 1} \frac{{x - 1}}{{(x - 1)\left( {\sqrt {x + 3} + 2} \right)}}\end{array}\]

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