C}{\partial U_n}=N(d_{1,n-1}).$$To verify these partials easily, one needs the identity$$\frac{\phi(d_{2i})}{\phi(d_{1i})}=\frac{U_{i+1}}{U_i}\text{ for }1\leq i\leq n-1,$$where $\phi(\bullet)$ is the standard normal density function. Observe that $U_i$ generally appears in the terms of (\[combo\]) that correspond to the indices $i-1, i, \text{and } i+1$. $U_1$ appears in the first two terms and $U_n$ appears in the last two terms. This being done, the delta-hedging strategy now obtains from the chain rule $$\frac{\partial C}{\partial S}=\sum\limits_{i=1}^n\frac{\partial C}{\partial U_i}\frac{\partial U_i}{\partial S}$$in conjunction with the fact that $\partial U_i/\partial S=b_iV_t(b_i)/S$. To get the horizon-free result, we just observe that $d_{1i}\to+\infty$ and $d_{2i}\to-\infty$ as $T\to\infty$. Finally, consider what happens when $\tau\to0$. The numbers $d_{1i},d_{2i}$ will converge to the same limit $\pm\infty$ according as $U_{i+1}$ is greater or less than $U_i$. In the event that $U_{i+1}=U_i$ then $d_{1i}$ and $d_{2i}$ both converge to zero. The numbers $(U_i)_{i=1}^n$ will typically have a unique mode $U_{i^*}$, e.g. $U_1<\cdot\cdot\cdot<U_{i^*-1}<U_{i^*}>U_{i^*+1}>\cdot\cdot\cdot>U_n$. In this case, all coefficients in the linear combination (\[combo\]) converge to zero except the one corresponding to $i=i^*$, which converges to $1$. If there are two modes $U_{i^*}=U_{i^*+1}$, then the corresponding coefficients in (\[combo\]) both converge to $1/2$, and the result follows.

Note that for $n>2$, the initial weighting of the $b_i$ 