Tính các giới hạn:

a) \(\left\{ \begin{array}{l}\mathop {\lim }\limits_{x \to {{12}^ + }} 2023 = 2023\\\mathop {\lim }\limits_{x \to {{12}^ + }} \left( {x - 12} \right) = 0;x - 12\rangle 0\end{array} \right. \Rightarrow \)\(\mathop {\lim }\limits_{x \to {{12}^ + }} \frac{{2023}}{{x - 12}} = + \infty \).

b) \(\mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {{x^2} + 4x} }}{{x - 1}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{ - x.\sqrt {1 + \frac{4}{x}} }}{{x\left( {1 - \frac{1}{x}} \right)}}\)\( = \mathop {\lim }\limits_{x \to - \infty } - \frac{{\sqrt {1 + \frac{4}{x}} }}{{1 - \frac{1}{x}}} = - 1\).