Answer
$$
\begin{aligned} \lim _{n \rightarrow \infty} &\left[\frac{R}{i-g}\left[1-\left(\frac{1+g}{1+i}\right)^{n}\right]\right] \\ &=\frac{R}{i-g} \lim _{n \rightarrow \infty}\left[1-\left(\frac{1+g}{1+i}\right)^{n}\right] \\ &=\frac{R}{i-g}\left[\lim _{n \rightarrow \infty} 1-\lim _{n \rightarrow \infty}\left(\frac{1+g}{1+i}\right)^{n}\right] \quad\quad\left[\begin{array}{c}{\text{ assuming }} \\ {i \gt g}\end{array}\right]\\
& =\frac{R}{i-g} \left[ 1-0\right]\\
&=\frac{R}{i-g}.
\end{aligned}
$$
Work Step by Step
$$
\frac{R}{i-g}\left[1-\left(\frac{1+g}{1+i}\right)^{n}\right]
$$
The limit of the expression above as $n$ approaches infinity is
$$
\begin{aligned} \lim _{n \rightarrow \infty} &\left[\frac{R}{i-g}\left[1-\left(\frac{1+g}{1+i}\right)^{n}\right]\right] \\ &=\frac{R}{i-g} \lim _{n \rightarrow \infty}\left[1-\left(\frac{1+g}{1+i}\right)^{n}\right] \\ &=\frac{R}{i-g}\left[\lim _{n \rightarrow \infty} 1-\lim _{n \rightarrow \infty}\left(\frac{1+g}{1+i}\right)^{n}\right] \quad\quad\left[\begin{array}{c}{\text{ assuming }} \\ {i \gt g}\end{array}\right]\\
& =\frac{R}{i-g} \left[ 1-0\right]\\
&=\frac{R}{i-g}.
\end{aligned}
$$
