Ta có
$\underset{x \to 1^+}{\lim} f(x) = \underset{x \to 1^+}{\lim} \dfrac{x^2 + x - 2}{x-1}$
$= \underset{x \to 1^+}{\lim} \dfrac{(x-1)(x+2)}{x-1}$
$= \underset{x \to 1^+}{\lim} x+2$
$= 3$
Lại có
$\underset{x \to 1^-}{\lim} f(x) = \underset{x \to 1^-}{\lim} x^2 + x + 1$
$= 3$
Ta có
$\underset{x \to 1^-}{\lim} f(x) = \underset{x \to 1^+}{\lim} f(x) (=3)$
Vậy hso liên tục tại $x = 1$.