This article is purely a matter of mathematical masturbation. Having solved an ODE with a continuous and finite difference method, we are going to show that
by taking the limit of the discrete solution as \(\Delta t \rightarrow\) 0 we can recover the continuous solution.

Recovering the continuous solution by limits

Let us review the ODE solutions we found and then show how to recover the continuous solution by taking a limit.
We began with an ordinary differential equation.

$$
\frac{\partial B}{\partial t} = Br - P
$$

The two solutions, discrete and continuous, look different but they should be equivalent in the limit where the discretization parameter \(\Delta t\) approaches 0.

Notice that the more complicated case is the same problem but wrapped in a definition of a definite integral.

$$
\lim_{\Delta t \rightarrow 0} \sum_{i=0}^{t/\Delta t-1} f(i) \Delta t = \int_0^{t}f(\tau) d\tau
$$

The limit of a sum is the sum of the limits.

$$
\begin{align}
\lim_{\Delta t \rightarrow 0} \sum_{i=0}^{t/\Delta t-1} & P \left(1+r\Delta t\right)^{i} \Delta t \\
\lim_{\Delta t \rightarrow 0} \sum_{i=0}^{t/\Delta t-1} & Pe^{rt} \Delta t\\
\int_0^{t} & Pe^{r\tau} d\tau\\
& \frac{P}{r}\left( e^{rt}-1\right)
\end{align}
$$

This section just serves to show there is no contradiction between the two methods - the limit of the discrete solution as \(\Delta t \rightarrow\) 0 is the continuous solution.

$$
\lim_{\Delta t \rightarrow 0} \left( B_0\left(1+r\Delta t\right)^{t/\Delta t} -\sum_{i=0}^{t/\Delta t-1} P \left(1+r\Delta t\right)^{i} \Delta t \right) = \left(B_0-\frac{P}{r}\right)e^{rt}+\frac{P}{r}
$$