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

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

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

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

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