Tính các giới hạn sau

                 a) \(\mathop {\lim }\limits_{x \to {3^ - }} \frac{{1 - 3x}}{{x - 3}}\)          ;           b) \[\mathop {\lim }\limits_{x \to 2} \frac{{3x - 6}}{{\sqrt {x + 2} - 2}}\]

a) Vì \(\mathop {\lim }\limits_{x \to {3^ - }} (1 - 3x) = 1 - 9 = - 8 < 0\) và \(\mathop {\lim }\limits_{x \to {3^ - }} (x - 3) = 0,\,\,\forall x < 3 \Rightarrow x - 3 < 0\)

\(\mathop {\lim }\limits_{x \to {3^ - }} \frac{{1 - 3x}}{{x - 3}} = + \infty \).

b) \[\mathop {\lim }\limits_{x \to 2} \frac{{3x - 6}}{{\sqrt {x + 2} - 2}} = \]\[\mathop {\lim }\limits_{x \to 2} \frac{{3(x - 2)\left( {\sqrt {x + 2} + 2} \right)}}{{\left( {\sqrt {x + 2} - 2} \right)\left( {\sqrt {x + 2} + 2} \right)}} = \mathop {\lim }\limits_{x \to 2} \frac{{3(x - 2)\left( {\sqrt {x + 2} + 2} \right)}}{{x + 2 - 4}}\]

\[ = \mathop {\lim }\limits_{x \to 2} 3\left( {\sqrt {x + 2} + 2} \right) = 12\].