If α & β be the roots of ax²+bx+c = 0, then lim_x→α( 1+ax²+bx+c)^1/(x-α) is :-

Hi

As α and β are roots

so,

ax² + bx + c = a(x - α) (x - β)

Let

y = lim_x->α {1+ a(x - α)(x - β)}^{1/(x - α)}

^=> log y = lim_x->α log [1+ a (x - α)(x - β)] / (x - α)

            = lim_x->α [a(x - α) + a (x - β)] / 1+a(x - β)(x-α)

             = a(α - β)

=> y = e^{a(α - β)}

^Thanks

Anurag Kishore