Đáp án:
$\begin{array}{l}
31)\mathop {\lim }\limits_{x \to {2^ + }} \left( {\dfrac{1}{{x - 2}} - \dfrac{1}{{{x^2} - 4}}} \right)\\
= \mathop {\lim }\limits_{x \to {2^ + }} \dfrac{{x + 2 - 1}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \mathop {\lim }\limits_{x \to {2^ + }} \dfrac{{x + 1}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{2 + 1}}{{\left( {2 - 2} \right).\left( {2 + 2} \right)}} = \dfrac{3}{{{0^ + }}} = + \infty \\
\Rightarrow B\\
32)\mathop {\lim }\limits_{x \to {1^ + }} \dfrac{{4x - 3}}{{x - 1}} = \dfrac{{4.1 - 3}}{{1 - 1}} = \dfrac{1}{{{0^ + }}} = + \infty \\
\Rightarrow C
\end{array}$
