1. Let $q(x)=x^2+bx+c.$ Given that $\alpha,\beta$ are distinct roots of $q(x)=0$, we have $q(\beta)=\beta^2+b\beta+c=0.$

2. The required limit is $L=\lim_{x\to \beta}\frac{e^{2q(x)}-1-2q(x)}{(x-\beta)^2}.$

3. Use the standard expansion near $t=0$: $e^{2t}=1+2t+\frac{(2t)^2}{2!}+\cdots=1+2t+2t^2+\cdots$ Hence, $e^{2t}-1-2t \sim 2t^2 \quad (t\to 0).$

Here, as $x\to\beta$, $q(x)\to q(\beta)=0$, so with $t=q(x)$, $e^{2q(x)}-1-2q(x) \sim 2(q(x))^2.$ Therefore, $L=\lim_{x\to\beta}2\left(\frac{q(x)}{x-\beta}\right)^2.$

4. Since $\beta$ is a root of $q(x)$, $q(x)=x^2+bx+c=(x-\alpha)(x-\beta).$ So, $\frac{q(x)}{x-\beta}=x-\alpha.$ Taking limit as $x\to\beta$, $\lim_{x\to\beta}\frac{q(x)}{x-\beta}=\beta-\alpha.$ Thus, $L=2(\beta-\alpha)^2.$

5. Now express $(\beta-\alpha)^2$ in terms of $b,c$. For the quadratic $x^2+bx+c=0$, $\alpha+\beta=-b, \qquad \alpha\beta=c.$ Hence, $(\beta-\alpha)^2=(\alpha+\beta)^2-4\alpha\beta=b^2-4c.$ Therefore, $L=2(b^2-4c).$

6. Compare with the options:

• A: $b^2+4c$
• B: $2(b^2+4c)$
• C: $2(b^2-4c)$
• D: $b^2-4c$

So the correct option is $\boxed{C}.$