If $\alpha$ and $\beta$ are the roots of the quadratic equation $ax^2 + bx + c = 0,$ then $\displaystyle\lim_{x\rightarrow \frac{1}{\alpha}}\sqrt{\dfrac{1-cos(cx^{2}+bx+a)}{2(1-\alpha x)^{2}}}$

Using the half-angle identity, the expression under the square root becomes $\sqrt{(1 - \cos(cx^2 + bx + a)) / (2(1 - \alpha x)^2)} = |\sin((cx^2 + bx + a) / 2) / (1 - \alpha x)|$. Because $1/\alpha$ is a root of $ax^2 + bx + c = 0$, the numerator evaluates to $0$ at $x = 1/\alpha$, allowing us to apply L'Hopital's rule to find the limit of the inner function. The limit becomes $|(c/\alpha + b) / 2| / |1 - \alpha(1/\alpha)| = |(c/\alpha + b) / 2| \cdot |1/\alpha - 1/\beta| / |1/\alpha - 1/\beta|$. Simplifying using the root relation gives the result $|c / (2\alpha) \cdot (1/\alpha - 1/\beta)|$.

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