Tính các giới hạn sau

a) \(\mathop {\lim }\limits_{n \to + \infty } \left( {\frac{{3n - 1}}{{2n + 3}}} \right) = \mathop {\lim }\limits_{n \to + \infty } \frac{{n\left( {3 - \frac{1}{n}} \right)}}{{n\left( {2 + \frac{3}{n}} \right)}}\)\( = \mathop {\lim }\limits_{n \to + \infty } \frac{{\left( {3 - \frac{1}{n}} \right)}}{{\left( {2 + \frac{3}{n}} \right)}} = \frac{3}{2}\)

Vì \(\mathop {\lim }\limits_{n \to + \infty } \frac{1}{n} = 0;\,\,\mathop {\lim }\limits_{n \to + \infty } \frac{3}{n} = 0\).

b) \(\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {2x + 1} - 1}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sqrt {2x + 1} - 1} \right)\left( {\sqrt {2x + 1} + 1} \right)}}{{x\left( {\sqrt {2x + 1} + 1} \right)}}\)

\( = \mathop {\lim }\limits_{x \to 0} \frac{{2x + 1 - 1}}{{x\left( {\sqrt {2x + 1} + 1} \right)}} = \mathop {\lim }\limits_{x \to 0} \frac{{2x}}{{x\left( {\sqrt {2x + 1} + 1} \right)}}\)

\( = \mathop {\lim }\limits_{x \to 0} \frac{2}{{\left( {\sqrt {2x + 1} + 1} \right)}} = \frac{2}{{\left( {\sqrt {2.0 + 1} + 1} \right)}} = 1.\)