Let \(\alpha(a)\) and \(\beta(a)\) be the roots of the equation \((\sqrt[3]{1+a}-1) x^2+(\sqrt{1+a}-1) x+(\sqrt[6]{1+a}-1)=0\) where \(a\gt-1\) then \(\lim _{a \rightarrow 0^{+}} \alpha(a)\) and \(\lim _{a \rightarrow 0^{+}} \beta(a)\) respectively are

Let \(A=1+a\)
\(\therefore \quad\) When \(\mathrm{a} \rightarrow 0^{+}, \mathrm{A} \rightarrow 1^{+}\)
\(\therefore \quad\) Given function is written as
\(\begin{aligned}
& \left(A^{\frac{1}{3}}-1\right) x^2+\left(A^{\frac{1}{2}}-1\right) x+\left(A^{\frac{1}{6}}-1\right)=0 \\
\therefore & \left(\frac{A^{\frac{1}{3}}-1}{A-1}\right) x^2+\left(\frac{A^{\frac{1}{2}}-1}{A-1}\right) x+\left(\frac{A^{\frac{1}{6}}-1}{A-1}\right)=0
\end{aligned}\)
Taking \(\lim _{x \rightarrow 0^{+}}\)on both sides, we get
\(\begin{array}{ll}
\therefore & \frac{1}{3} x^2+\frac{1}{2} x+\frac{1}{6}=0 \\
\therefore & 2 x^2+3 x+1=0 \\
\therefore & x=-1 \text { or } \frac{-1}{2}
\end{array}\)
i.e. \(\lim _{x \rightarrow 0^{+}} \alpha(a)=-1\) and \(\lim _{x \rightarrow 0^{+}} \beta(a)=\frac{-1}{2}\)