Tính các giới hạn sau:a) (mathop {lim }limits_{x o  - infty } left( {sqrt {9{x^2} + 12x}  + 3x} ight))b) (mathop {l

a) \(\mathop {\lim }\limits_{x \to  - \infty } \left( {\sqrt {9{x^2} + 12x}  + 3x} \right) = \mathop {\lim }\limits_{x \to  - \infty } \left( {\frac{{12x}}{{\sqrt {9{x^2} + 12x}  - 3x}}} \right)\)

\( = \mathop {\lim }\limits_{x \to  - \infty } \left( {\frac{{12}}{{ - \sqrt {9 + \frac{{12}}{{{x^2}}}}  - 3}}} \right) = 2\)

b) \(\mathop {\lim }\limits_{x \to {3^ + }} \frac{{\left| {3 - x} \right|\sqrt {{x^2} + 7}  - 4\left( {x - 3} \right)}}{{{{\left( {x - 3} \right)}^2}}} = \mathop {\lim }\limits_{x \to {3^ + }} \frac{{\sqrt {{x^2} + 7}  - 4}}{{\left( {x - 3} \right)}}\)

\(\begin{array}{l}
 = \mathop {\lim }\limits_{x \to {3^ + }} \frac{{x + 3}}{{\sqrt {{x^2} + 7}  + 4}}\\
 = \frac{3}{4}
\end{array}\)