Tính các giới hạn sau

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

\[ = \mathop {\lim }\limits_{n \to + \infty } \frac{{3n}}{{n\sqrt {1 + \frac{3}{n}} + n}} = \mathop {\lim }\limits_{n \to + \infty } \frac{3}{{\sqrt {1 + \frac{3}{n}} + 1}} = \frac{3}{2}.\]

b) Ta có: \[\mathop {\lim }\limits_{x \to 2} \left( {5 - x} \right) = 5 - 2 = 3 > 0;\]

\[\mathop {\lim }\limits_{x \to 2} {\left( {x - 2} \right)^2} = 0,\,\,{\left( {x - 2} \right)^2} > 0\,\,\forall x \ne 2.\]

Do đó, \[\mathop {\lim }\limits_{x \to 2} \frac{{5 - x}}{{{{\left( {x - 2} \right)}^2}}} = + \infty .\]