Tính các giới hạn sau:
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Tính các giới hạn sau:
a) \(\mathop {\lim }\limits_{n \to + \infty } \frac{{3n - 1}}{{2n + 3}}.\) b) . 
a)\(\lim \frac{{3n - 1}}{{2n + 3}} = \lim \frac{{n\left( {3 - \frac{1}{n}} \right)}}{{n\left( {2 + \frac{3}{n}} \right)}} = \lim \frac{{\left( {3 - \frac{1}{n}} \right)}}{{\left( {2 + \frac{3}{n}} \right)}} = \frac{3}{2}\).
b) \(\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {2x + 1} - 1}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sqrt {2x + 1} - 1} \right)\left( {\sqrt {2x + 1} + 1} \right)}}{{x\left( {\sqrt {2x + 1} + 1} \right)}}\)
\( = \mathop {\lim }\limits_{x \to 0} \frac{{2x + 1 - 1}}{{x\left( {\sqrt {2x + 1} + 1} \right)}} = \mathop {\lim }\limits_{x \to 0} \frac{{2x}}{{x\left( {\sqrt {2x + 1} + 1} \right)}}\)
\( = \mathop {\lim }\limits_{x \to 0} \frac{2}{{\left( {\sqrt {2x + 1} + 1} \right)}} = \frac{2}{{\left( {\sqrt {2.0 + 1} + 1} \right)}} = 1\).