Limits of Roots of Quadratic Equation as Parameter Approaches Zero

To find the limiting values of the roots $\alpha(a)$ and $\beta(a)$ as $a \to 0^+$, we analyze the given quadratic equation:

$(\sqrt[3]{1+a}-1)x^2+(\sqrt{1+a}-1)x+(\sqrt[6]{1+a}-1)=0$

For $a > 0$, we divide the entire equation by $a$:

$\left( \frac{(1+a)^{1/3}-1}{a} \right) x^2 + \left( \frac{(1+a)^{1/2}-1}{a} \right) x + \left( \frac{(1+a)^{1/6}-1}{a} \right) = 0$

Recall the standard limit formula derived from the binomial expansion or L'Hôpital's rule: $\lim_{a \to 0} \frac{(1+a)^n - 1}{a} = n$

Using this standard limit, we evaluate the limit of each coefficient as $a \to 0^+$:

1. Coefficient of $x^2$: $\lim_{a \to 0^+} \frac{(1+a)^{1/3}-1}{a} = \frac{1}{3}$

2. Coefficient of $x$: $\lim_{a \to 0^+} \frac{(1+a)^{1/2}-1}{a} = \frac{1}{2}$

3. Constant term: $\lim_{a \to 0^+} \frac{(1+a)^{1/6}-1}{a} = \frac{1}{6}$

Since the coefficient of $x^2$ tends to a non-zero finite value ($\frac{1}{3} \neq 0$), the roots of the quadratic equation continuous in $a$ will converge to the roots of the limiting quadratic equation:

$\frac{1}{3}x^2 + \frac{1}{2}x + \frac{1}{6} = 0$

Multiplying the entire equation by $6$ to clear the fractions:

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

Factoring the quadratic expression:

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

Thus, the roots of the limiting equation are:

$x = -\frac{1}{2} \quad \text{and} \quad x = -1$

Therefore:

$\lim_{a \to 0^+} \alpha(a) = -\frac{1}{2} \quad \text{and} \quad \lim_{a \to 0^+} \beta(a) = -1$

This matches Option (B).