Ta có
$\underset{x \to 1}{\lim} \dfrac{\sqrt{x^2 + 2} - \sqrt{4 - x}}{x-1} = \underset{x \to 1}{\lim} \dfrac{\sqrt{x^2 +2} - \sqrt{3} - (\sqrt{4-x} - \sqrt{3})}{x-1}$
$= \underset{x \to 1}{\lim} \dfrac{\frac{x^2 + 2 - 3}{\sqrt{x^2 + 2} + \sqrt{3}} - \frac{4 - x - 3}{\sqrt{4-x} + \sqrt{3}}}{x-1}$
$= \underset{x \to 1}{\lim} \dfrac{\frac{(x-1)(x+1)}{\sqrt{x^2 + 2} + \sqrt{3}} - \frac{1-x}{\sqrt{4-x} + \sqrt{3}}}{x-1}$
$= \underset{x \to 1}{\lim} \dfrac{x+1}{\sqrt{x^2 + 2} + \sqrt{3}} + \dfrac{1}{\sqrt{4-x} + \sqrt{3}}$
$= \dfrac{2}{\sqrt{3} + \sqrt{3}} + \dfrac{1}{\sqrt{3} + \sqrt{3}}$
$= \dfrac{\sqrt{3}}{2}$
Vậy $a = 3, b = 2$.
Suy ra $a + b = 3 + 2 = 5$.
