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Q. If $\alpha, \beta$ are the distinct roots of $x ^{2}+ bx + c =0$ then $\displaystyle\lim _{x \rightarrow \beta} \frac{e^{2\left(x^{2}+b x+c\right)}-1-2\left(x^{2}+b x+c\right)}{(x-\beta)^{2}}$ is equal to:

JEE MainJEE Main 2021Limits and Derivatives

Solution:

$\displaystyle\lim _{x \rightarrow \beta} \frac{e^{2\left(x^{2}+b x+c\right)}-1-2\left(x^{2}+b x+c\right)}{(x-\beta)^{2}}$
$\Rightarrow \displaystyle\lim _{x \rightarrow \beta} \frac{1\left(1+\frac{2\left(x^{2}+b x+c\right)}{1 !}+\frac{2^{2}\left(x^{2}+b x+c\right)^{2}}{2 !}+\ldots\right)-1-2\left(x^{2}+b x+c\right)}{(x-\beta)^{2}}$
$\Rightarrow \displaystyle\lim _{x \rightarrow \beta} \frac{2\left(x^{2}+b x+1\right)^{2}}{(x-\beta)^{2}}$
$\Rightarrow \displaystyle\lim _{x \rightarrow \beta} \frac{2(x-\alpha)^{2}(x-\beta)^{2}}{(x-\beta)^{2}}$
$\Rightarrow 2(\beta-\alpha)^{2}=2\left(b^{2}-4 c\right)$