Ta tính
$\underset{x \to -1}{\lim} f(x) = \underset{x \to -1}{\lim} \dfrac{2x + \sqrt{x^2 + x + 4}}{x+1}$
$= \underset{x \to -1}{\lim} \dfrac{4x^2 - (x^2 + x + 4)}{(x+1)(2x - \sqrt{x^2 + x +4})}$
$= \underset{x \to -1}{\lim} \dfrac{3x^2 - x - 4}{(x+1)(2x - \sqrt{x^2 + x + 4})}$
$= \underset{x \to -1}{\lim} \dfrac{(x+1)(3x-4)}{(x+1)(2x - \sqrt{x^2 + x + 4})}$
$= \underset{x \to -1}{\lim} \dfrac{3x-4}{2x - \sqrt{x^2 + x + 4}}$
$= \dfrac{-3-4}{-2-2} = \dfrac{7}{4}$
Để hso liên tục tại $x_0 = -1$ thì ta phải có
$\underset{x \to -1}{\lim} f(x) = f(-1)$
$<-> \dfrac{7}{4} = (-a)^2 + a(-1) + \dfrac{7}{4}$
$<-> a^2 - a = 0$
$<-> a(a-1) = 0$
Vậy $a = 0$ hoặc $a = 1$.
