a) \(x \to {3^ + } \Rightarrow x - 3 > 0\)
\(\mathop {\lim }\limits_{x \to {3^ + }} \frac{{{x^2} - 9}}{{\left| {x - 3} \right|}} = \mathop {\lim }\limits_{x \to {3^ + }} \frac{{{x^2} - 9}}{{x - 3}} = \mathop {\lim }\limits_{x \to {3^ + }} \left( {x + 3} \right) = 6\)
b) \(\mathop {\lim }\limits_{x \to {1^ - }} x = 1\)
\(\mathop {\lim }\limits_{x \to {1^ - }} \frac{1}{{\sqrt {1 - x} }} = 0 \)
Mà \(x \to {1^ - } \Rightarrow 1 - x > 0 \Rightarrow \sqrt {1 - x} > 0\)
Suy ra \( \mathop {\lim }\limits_{x \to {1^ - }} \frac{x}{{\sqrt {1 - x} }} = + \infty \)
