Ta có: \(f\left( 1 \right) = 4\)
\(\mathop {\lim }\limits_{x \to 1} f\left( x \right) = \mathop {\lim }\limits_{x \to 1} \dfrac{{{x^2} + 2x - 3}}{{x - 1}}\) \( = \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {x - 1} \right)\left( {x + 3} \right)}}{{x - 1}}\) \( = \mathop {\lim }\limits_{x \to 1} \left( {x + 3} \right) = 4\)
Do đó \(\mathop {\lim }\limits_{x \to 1} f\left( x \right) = f\left( 1 \right) = 4\) nên hàm số liên tục tại \(x = 1\).