Problem source

The value $V_N$ of a bond after $N$ days is determined by the equation
$$
V_{N+1} = (1+c) V_{N} -d \qquad (c>0, \ d>0),
$$
where $c$ and $d$ are given  constants. 
By looking for solutions of the form $V_T= A k^T + B$ for some constants $A,B$ and $k$, or otherwise, find $V_N$ in terms of $V_0$. 
What is the solution for $c=0$? Show that this is the limit  (for fixed $N$) as $c\rightarrow 0$ of your solution for $c>0$.

Solution source

Suppose $V_T = Ak^T + B$, then

\begin{align*}
&& Ak^{T+1}+B &= (1+c)(Ak^T+B) - d \\
\Rightarrow && (k-1-c)Ak^T &= cB -d \\
\Rightarrow && k &= 1+c \\
&& B &= \frac{d}{c} \\
&& A &= V_0 - B \\
\Rightarrow && V_N &= (V_0 - \frac{d}{c})(1+c)^{N} + \frac{d}{c}
\end{align*}

When $c = 0$, $V_{N+1} = V_N - d \Rightarrow V_N = V_0 - Nd$.

\begin{align*}
\lim_{c \to 0} \left ( (V_0 - \frac{d}{c})(1+c)^{N} + \frac{d}{c} \right) &= \lim_{c \to 0}  \left ( \frac{(V_0 c-d)(1+c)^N + d}{c} \right ) \\
&= \lim_{c \to 0}  \left ( \frac{V_0c - d-Ncd+NV_0c^2 + o(c^2) + d}{c} \right ) \\
&= \lim_{c \to 0}  \left ( V_0 - Nd + o(c) \right ) \\
&= V_0 - Nd
\end{align*}