$\displaystyle\lim_{x \to 1^-} f(x)=\displaystyle\lim_{x \to 1^-} (2x+a)=2+a\\ \displaystyle\lim_{x \to 1^+} f(x)=\displaystyle\lim_{x \to 1^+} \dfrac{x^3-x^2+2x-2}{x-1}\\ =\displaystyle\lim_{x \to 1^+} \dfrac{x^2(x-1)+2(x-1)}{x-1}\\ =\displaystyle\lim_{x \to 1^+} (x^2+2)\\ =3$
Để hàm số liên tục trên $\mathbb{R}$
$\displaystyle\lim_{x \to 1^-} f(x)=\displaystyle\lim_{x \to 1^+} f(x)\\ \Leftrightarrow 2+a=3\\ \Leftrightarrow a=1$
