Question

Let $\alpha$ and $\beta$ be the distinct roots of $ax^2 + bx + c = 0$ , then $\displaystyle \lim_{x \to\alpha} \frac{1- \cos \left(ax^{2} + bx + c\right)}{\left(x-\alpha\right)^{2}} $ is equal to

• A. $\frac{a^2}{2} (\alpha - \beta)^2$
• B. 0
• C. $\frac{-a^2}{2} (\alpha - \beta )^2$
• D. $\frac{1}{2} (\alpha - \beta )^2$

Answer 1

Answer: (A)

Given limit $= \displaystyle \lim_{x \to\alpha} \frac{1- \cos a\left(x -\alpha\right)\left(x-\beta\right)}{\left(x -\alpha\right)^{2}} $
$= \displaystyle \lim_{x \to\alpha} \frac{2\sin^{2} \left(a \frac{\left(x -\alpha\right)\left(x - \beta\right)}{2}\right)}{\left(x -\alpha\right)^{2}} $
$=\displaystyle \lim_{x \to\alpha } \frac{2}{\left(x -\alpha\right)^{2}} \times\frac{\sin^{2}\left(a \frac{\left(x-\alpha\right)\left(x -\beta\right)}{2}\right)}{\frac{a^{2}\left(x-\alpha\right)^{2}\left(x-\beta\right)^{2}}{4}} \times\frac{a^{2} \left(x-\alpha\right)^{2} \left(x -\beta\right)^{2}}{4}$
$ = \frac{a^{2}\left(\alpha - \beta\right)^{2}}{2} $.