I would start by letting and then take the log of both sides and apply the property of logarithms to write it like this:
And note that you can take the limit of both sides and solve for the limit of as
EDIT, typo:
I would start by letting and then take the log of both sides and apply the property of logarithms to write it like this:
And note that you can take the limit of both sides and solve for the limit of as
EDIT, typo:
Got it,
Rewrite as Ln y= Ln[x(x-1)T+1] / (1-x) then apply l'hopitals,
Rewrite Ln[x(x-1)T+1]=a (1-x)=b
Lim X->1 =da/dx / db/dx
da/dx = T(2x-1)/[Tx(x-1)+1]
db/dx = -1
da/dx / db/dx = -T(2x-1)/[Tx(x-1)+1] = -T
ln(y) = -T
Y= e^-T
Sorry for many edits !