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Q. Let $\alpha$ and $\beta$ be the distinct roots of $a x^{2}+b x+c=0$, then $\displaystyle\lim _{x \rightarrow \alpha} \frac{1-\cos \left(a x^{2}+b x+c\right)}{(x-\alpha)^{2}}$ is equal to

Solution:

Now, $\displaystyle\lim _{x \rightarrow \alpha} \frac{1-\cos \left(a x^{2}+b x+c\right)}{(x-\alpha)^{2}}$
$=\displaystyle\lim _{x \rightarrow \alpha} \frac{2 \sin ^{2}\left(\frac{a x^{2}+b x+c}{2}\right)}{(x-\alpha)^{2}}$
$=\displaystyle\lim _{x \rightarrow \alpha} \frac{2 \sin ^{2}\left(\frac{a}{2}(x-\alpha)(x-\beta)\right)}{\left(\frac{a}{2}\right)^{2}(x-\alpha)^{2}(x-\beta)^{2}}\left(\frac{a}{2}\right)^{2}(x-\beta)^{2}$
$=\displaystyle\lim _{x \rightarrow \alpha} \frac{a^{2}}{2}(x-\beta)^{2}\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]$
$=\frac{a^{2}}{2}(\alpha-\beta)^{2}$