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Mathematics Question on limits and derivatives

Let α\alpha and β\beta be the distinct roots of ax2+bx+c=0ax^2 + bx + c = 0 , then lim⁡x→α1−cos⁡(ax2+bx+c)(x−α)2\displaystyle \lim_{x \to\alpha} \frac{1- \cos \left(ax^{2} + bx + c\right)}{\left(x-\alpha\right)^{2}} is equal to

A

a22(α−β)2\frac{a^2}{2} (\alpha - \beta)^2

B

0

C

−a22(α−β)2\frac{-a^2}{2} (\alpha - \beta )^2

D

12(α−β)2\frac{1}{2} (\alpha - \beta )^2

Answer

a22(α−β)2\frac{a^2}{2} (\alpha - \beta)^2

Explanation

Solution

Given limit =lim⁡x→α1−cos⁡a(x−α)(x−β)(x−α)2= \displaystyle \lim_{x \to\alpha} \frac{1- \cos a\left(x -\alpha\right)\left(x-\beta\right)}{\left(x -\alpha\right)^{2}} =lim⁡x→α2sin⁡2(a(x−α)(x−β)2)(x−α)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}} =lim⁡x→α2(x−α)2×sin⁡2(a(x−α)(x−β)2)a2(x−α)2(x−β)24×a2(x−α)2(x−β)24=\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} =a2(α−β)22= \frac{a^{2}\left(\alpha - \beta\right)^{2}}{2}.