Tính các giới hạn sau:

a) \(\lim \frac{{\sqrt {4{n^2} + 1} - \sqrt {n + 2} }}{{2n - 3}}\)\( = \lim \frac{{\sqrt {4 + \frac{1}{{{n^2}}}} - \sqrt {\frac{1}{n} + \frac{2}{{{n^2}}}} }}{{2 - \frac{3}{n}}}\)

\( = \frac{{\lim \left( {\sqrt {4 + \frac{1}{{{n^2}}}} - \sqrt {\frac{1}{n} + \frac{2}{{{n^2}}}} } \right)}}{{\lim \left( {2 - \frac{3}{n}} \right)}} = \frac{{\lim \sqrt {4 + \frac{1}{{{n^2}}}} - \lim \sqrt {\frac{1}{n} + \frac{2}{{{n^2}}}} }}{{\lim \left( {2 - \frac{3}{n}} \right)}} = \frac{{\sqrt 4 - \sqrt 0 }}{2} = 1.\)

b) \(\mathop {\lim }\limits_{x \to {1^ + }} \frac{{\sqrt {{x^3} - {x^2}} }}{{\sqrt {x - 1} + 1 - x}}\)\( = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{\sqrt {{x^2}\left( {x - 1} \right)} }}{{\sqrt {x - 1} + 1 - x}}\)\( = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{x\sqrt {\left( {x - 1} \right)} }}{{\sqrt {x - 1} \left( {1 - \sqrt {x - 1} } \right)}}\)

\( = \mathop {\lim }\limits_{x \to {1^ + }} \frac{x}{{1 - \sqrt {x - 1} }} = \frac{1}{{1 - \sqrt {1 - 1} }} = 1.\)