Asymptotic behavior of European options [closed]

I guess $x$ is the log moneyness $\log(S_0/K)$ where $S_0$ is today's stock price. Then $\lim\limits_{x\to-\infty}$ corresponds to $\lim\limits_{S_0\to 0}$ and $\lim\limits_{x\to+\infty}$ corresponds to $\lim\limits_{S_0\to+\infty}\,.$ It is clear that in any model for the dynamics of $S_t$ we must have (abusing notation) $$ \lim_{S_0\to 0}u(\tau,S_0)=0\, $$ because a call option on a stock must be worth zero if the stock goes to zero. This is the first boundary condition. For very large $S_0$ (or $x$) we must have in any model \begin{align} u(\tau,S_0)&\approx S_0\,e^{-\int_0^\tau D(s)\,ds}-Ke^{-\int_0^\tau r(s)\,ds}\\ &=K\left(e^{x-\int_0^\tau D(s)\,ds}-e^{-\int_0^\tau r(s)\,ds} \right)\,, \end{align} because for a very large stock price the call option must approach that difference regardless of the stock volatility, or the dynamics of $S_t$. This is the second boundary condition.

In a tree or PDE approach you choose a large enough upper boundary at which you apply the second boundary condition. You ignore all values of $S_t$ that are above that boundary. The hedging strategy was only a no arbitrage argument used to derive the pricing equation that you are solving. I don't think you need to worry about the hedging strategy at those boundaries, unless I misunderstood question 2.