If α, β are the distinct roots of x² + bx + c = 0, then $\underset{x\to \beta }{lim}\frac{e^{2\left(x^2+bx+c\right)}-1-2\left(x^2+bx+c\right)}{{\left(x-\beta \right)}^2}$  is equal to

Given let α, β be the roots of the equation x² + bx + c = 0

So, ${\alpha }^2+b\alpha +c=0\&{\beta }^2+b\beta +c=0$

Also x² + bx + c = (x - α) (x - β)

$NowL=\underset{x\to \beta }{lim}\frac{e^{2\left(x^2+bx+c\right)}-1-2\left(x^2+bx+c\right)}{{\left(x-\beta \right)}^2}$           

$\underset{x\to \beta }{lim}\frac{2{\left(x-\alpha \right)}^2{\left(x-\beta \right)}^2+\frac{8}{6}{\left(x-\alpha \right)}^3{\left(x-\beta \right)}^3+........}{{\left(x-\beta \right)}^2}$

=  $2{\left(\beta -\alpha \right)}^2=2\left[{\left(\beta +\alpha \right)}^2-4\alpha \beta \right]=2\left[b^2-4c\right]$