We need to show the following,
Lim X-> 1
[X(X-1)T+1]^(1/(1-X)) = EXP(-T)
Obviously we need to apply L'Hopitals rule, but it is not straight forward how to proceed !
I would start by letting $\displaystyle y=[x(x-1)T+1]^{\frac{1}{1-x}}$ and then take the log of both sides and apply the property of logarithms to write it like this:
$\displaystyle ln(y)=\frac{1}{1-x}\ln[x(x-1)T+1]$
And note that you can take the limit of both sides and solve for the limit of $\displaystyle y$ as $\displaystyle x->1$
EDIT, typo:
$\displaystyle \lim_{x->1}ln(y)=\lim_{x->1}\frac{\ln[x(x-1)T+1]}{1-x}$
ok, note that I had to edit my post. Now, you know how to deal with the right hand side right? You mentioned L-hopitals rule which sounds right to me since it has the forum $\displaystyle \frac{0}{0}$ at $\displaystyle x=1$ so try that, and you can solve the equation for $\displaystyle \lim_{x->1}y$.
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 !