Mathematics - Limits Question with Solution | TestHub

$(\sqrt[3]{1+a}-1)x^2+(\sqrt{1+a}-1)x+(\sqrt[6]{1+a}-1)=0$ 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 $y\to 1$ i.e. $a\to 0$ on both sides, we get $\frac{1}{3}{x}^2+\frac{1}{2}x+\frac{1}{6}=0\Rightarrow 2x^2+3x+1=0$ $\Rightarrow x=-1,-\frac{1}{2}$ (roots of the equation) $\therefore {\lim }_{a\to 0^{+}}\alpha (a)=-1,{\lim }_{a\to 0^{+}}\beta (a)=-\frac{1}{2}$