Let a + 1 = t⁶. Thus, when a $$\to$$ 0, t $$\to$$ 1.

$$\therefore$$ $$({t^2} - 1){x^2} + ({t^3} - 1)x + (t - 1) = 0$$

$$ \Rightarrow (t - 1)\{ (t + 1){x^2} + ({t^2} + t + 1)x + 1\} = 0$$,

as t $$\to$$ 1

$$2{x^2} + 3x + 1 = 0$$

$$ \Rightarrow 2{x^2} + 2x + x + 1 = 0$$

$$ \Rightarrow (2x + 1)(x + 1) = 0$$

Thus, x = $$-$$1, $$-$$1/2

or, $$\mathop {\lim }\limits_{a \to {0^ + }} \alpha (a) = - {1 \over 2}$$

and $$\mathop {\lim }\limits_{a \to {0^ + }} \beta (a) = - 1$$