Khi \(x \to {2^ - } \Rightarrow \left| {x - 2} \right| = 2 - x\)
Ta có:
\(\mathop {\lim }\limits_{x \to {2^ - }} \frac{{{x^2} - 5x + 6}}{{\left| {x - 2} \right|}} = \mathop {\lim }\limits_{x \to {2^ - }} \frac{{{x^2} - 5x + 6}}{{2 - x}} = \mathop {\lim }\limits_{x \to {2^ - }} \frac{{\left( {x - 2} \right)\left( {x - 3} \right)}}{{ - \left( {x - 2} \right)}}= \mathop {\lim }\limits_{x \to {2^ - }} \left[ { - \left( {x - 3} \right)} \right] = 3 - 2 = 1\)
Khi \(x \to {2^ + } \Rightarrow \left| {x - 2} \right| = x - 2\)
Ta có
\(\mathop {\lim }\limits_{x \to {2^ + }} \frac{{{x^2} - 5x + 6}}{{\left| {x - 2} \right|}} = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{{x^2} - 5x + 6}}{{x - 2}} = \mathop {\lim }\limits_{x \to {2^ + }} \frac{{\left( {x - 2} \right)\left( {x - 3} \right)}}{{x - 2}} = \mathop {\lim }\limits_{x \to {2^ - }} \left( {x - 3} \right) = 2 - 3 = - 1\)
