About the boundary conditions of the Black-Scholes-Merton PDE I have a question about the solution of the Black-Scholes PDE for the European call option when I read the book Stochastic Calculus for Finance II of Steven E.Shreve.
Let $c(t,x)$ be the value of the European call option at time $t$ if the stock price at that time is $S(t)=x$. Then, $c(t,x)$ satisfies the following equation:
$$c_t(t,x) + r x c_x(t,x)+ \frac{1}{2} \sigma^2 x^2 c_{xx}(t,x)=rc(t,x) \text{ for all $t\in [0, T), x\geq 0$},$$
and $c(T, x)=(x-K)^{+}.$
To resolve the above equation, one needs boundary conditions at $x=0$ and $x= +\infty$. For $x=0$. It's easy to derive that $c(t,0)=0$ for all $t \in [0,T]$.
However, for $x=+\infty$, I do not know understand how the author finds out (w/o a detailed explanation)(c.f. page 158 in that book) that
$$\lim_{x \rightarrow +\infty} c(t,x)- (x- e^{-r (T-t)}K) =0 \text{ for all $t \in [0, T]$}?$$

lulu's comment sums it up nicely.  Shreve mentions this is a backward parabolic PDE, and for it to be well-defined we must specify boundary conditions.  From a purely mathematical viewpoint, those can be almost anything we want.  The condition $$\lim_{x \rightarrow +\infty} c(t,x)- (x- e^{-r (T-t) K}) =0 \text{ for all $t \in [0, T]$}$$ is purely an economic argument, as pointed out by lulu.
