Let α(a) and β(a) be the roots of the equation ( [3] 1+a -1) x^ 2...

$(\sqrt[3]{1+a}-1) x^{2}+(\sqrt{1+a}-1) x+(\sqrt[6]{1+a}-1)=0$ Let $a+1=y$, then equation reduces to \left(y^{1 / 3}-1\right) x^{2}+$\left(y^{1 / 2}-1\right) x+\left(y^{1 / 6}-1\right)=0$ On dividing both sides by $y-1$, we get \left(\frac{y^{1 / 3}-1}{y-1}\right) x^{2}+$\left(\frac{y^{1 / 2}-1}{y-1}\right) x+\left(\frac{y^{1 / 6}-1}{y-1}\right)=0$ On taking limit as $y \rightarrow 1$ i.e. $a \rightarrow 0$ on both sides, we get $\frac{1}{3} x^{2}+\frac{1}{2} x+\frac{1}{6}=0 \Rightarrow 2 x^{2}+3 x+1=0 \Rightarrow x=-1,-\frac{1}{2}$ (roots of the equation) $\therefore \quad \lim _{a \rightarrow 0^{+}} \alpha(a)=-1, \lim _{a \rightarrow 0^{+}} \beta(a)=-\frac{1}{2}$