Consider the scattering of a quantum particle in one dimension, caused by a step in the potential (this appears in many undergrad level QM books):
$$ V(x) = \begin{cases} V_1 & x<0 \\ V_2 & x>0\end{cases}. $$
The particle is incident from the left, so it's wavefunction is:
$$ \psi(x) = \begin{cases} e^{i k_1 x} + r e^{-i k_1 x} & x<0 \\ t e^{i k_2 x} & x>0\end{cases}, $$
where $k_i =\sqrt{2m(E-V_i)}/\hbar$.
Matching the wavefunction and its derivative at $x=0$ gives:
$$ r = \frac{k_1-k_2}{k_1+k_2} ~~~;~~~ t = \frac{2 k_1}{k_1+k_2}.$$
Now we put another step in the potential at some distance $L$, which makes it a box potential:
$$ V(x) = \begin{cases} V_1 & x<0 \\ V_2 & 0<x<L \\ V_1 & L<x\end{cases}. $$
We solve this in a similar manner as before, with the wavefunction:
$$ \psi(x) = \begin{cases} e^{i k_1 x} + r e^{-i k_1 x} & x<0 \\ a e^{i k_2 x} + b e^{-i k_2 x} & 0<x<L \\ t e^{i k_1 x} & L<x \end{cases}. $$
Matching the wavefunction and its derivative at $x=0,L$ gives:
$$ r = \frac{k_1^2-k_2^2}{k_1^2+k_2^2+2 i k_1 k_2 \cot{(k_2 L)} } ~~~;~~~ t = \text{(something)}.$$
How come the second scattering problem doesn't reproduce the first scattering problem in the limit $L \rightarrow \infty$?
I'm looking only at the value of $r$. I send a particle in, it scatters, and I get something back with an amplitude $r$. It seems unphysical that if the potential changed at $x=L$, it changes the scattering at $x=0$, no matter how far $L$ is.