If alpha, beta are the distinct roots of x^{2 | vedclass

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(C) Let $f(x) = x^{2}+bx+c$. Since $\alpha, \beta$ are roots,$f(x) = (x-\alpha)(x-\beta)$.As $x \rightarrow \beta$,$f(x) \rightarrow 0$.Using the Tayl

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$2(b^{2}-4c)$

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(C) Let $f(x) = x^{2}+bx+c$. Since $\alpha, \beta$ are roots,$f(x) = (x-\alpha)(x-\beta)$.As $x \rightarrow \beta$,$f(x) \rightarrow 0$.Using the Taylor series expansion $e^{u} = 1 + u + \frac{u^{2}}{2!} + \dots$,where $u = 2f(x)$:$\lim _{x \rightarrow \beta} \frac{e^{2f(x)}-1-2f(x)}{(x-\beta)^{2}} = \lim _{x \rightarrow \beta} \frac{(1 + 2f(x) + \frac{(2f(x))^{2}}{2} + \dots) - 1 - 2f(x)}{(x-\beta)^{2}}$$= \lim _{x \rightarrow \beta} \frac{2(f(x))^{2}}{(x-\beta)^{2}}$$= \lim _{x \rightarrow \beta} \frac{2((x-\alpha)(x-\beta))^{2}}{(x-\beta)^{2}}$$= \lim _{x \rightarrow \beta} 2(x-\alpha)^{2} = 2(\beta-\alpha)^{2}$Since $(\beta-\alpha)^{2} = (\beta+\alpha)^{2} - 4\alpha\beta = (-b)^{2} - 4c = b^{2}-4c$,The limit is $2(b^{2}-4c)$.

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

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