Ta có:
\(\eqalign{
& \mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} \left( {mx + m + 1} \right) = 3m + 1 = f\left( 2 \right) \cr
& \mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ - }} {{{x^2} - 3x + 2} \over {{x^2} - 2x}}\cr& = \mathop {\lim }\limits_{x \to {2^ - }} {{\left( {x - 1} \right)\left( {x - 2} \right)} \over {x\left( {x - 2} \right)}} = \mathop {\lim }\limits_{x \to {2^ - }} {{x - 1} \over x} = {1 \over 2} \cr} \)
f liên tục tại mọi \(x ≠ 2\). Do đó :
f liên tục trên \(\mathbb R ⇔\) f liên tục tại \(x = 2\)
\(⇔ \mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = f\left( 2 \right) \)
\(\Leftrightarrow 3m + 1 = {1 \over 2} \Leftrightarrow m = - {1 \over 6}\)
