Let alpha(a) and beta(a) be the roots of the | vedclass

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(B) Let $A = 1 + a$. As $a \rightarrow 0^{+}$,$A \rightarrow 1^{+}$. The given equation is $(A^{\frac{1}{3}}-1) x^2+(A^{\frac{1}{2}}-1) x+(A^{\frac{1}

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$-1$ and $-\frac{1}{2}$

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(B) Let $A = 1 + a$. As $a ightarrow 0^{+}$,$A ightarrow 1^{+}$. The given equation is $(A^{\frac{1}{3}}-1) x^2+(A^{\frac{1}{2}}-1) x+(A^{\frac{1}{6}}-1)=0$. Dividing the entire equation by $(A-1)$ where $A 
eq 1$: $\frac{A^{\frac{1}{3}}-1}{A-1} x^2 + \frac{A^{\frac{1}{2}}-1}{A-1} x + \frac{A^{\frac{1}{6}}-1}{A-1} = 0$. Taking the limit as $A ightarrow 1$ (which corresponds to $a ightarrow 0^{+}$),we use the standard limit $\lim _{A ightarrow 1} \frac{A^n-1}{A-1} = n$: $\frac{1}{3} x^2 + \frac{1}{2} x + \frac{1}{6} = 0$. Multiplying by $6$ to clear denominators: $2 x^2 + 3 x + 1 = 0$. Factoring the quadratic: $(2x + 1)(x + 1) = 0$. Thus,the roots are $x = -1$ and $x = -\frac{1}{2}$. Therefore,$\lim _{a ightarrow 0^{+}} \alpha(a) = -1$ and $\lim _{a ightarrow 0^{+}} \beta(a) = -\frac{1}{2}$.

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 $a > -1$. Then $\lim _{a \rightarrow 0^{+}} \alpha(a)$ and $\lim _{a \rightarrow 0^{+}} \beta(a)$ respectively are

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