Answer
$$ \approx 1094.1742$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{n \to \infty } 1000{\left( {1 + \frac{{0.09}}{n}} \right)^n} \cr
& {\text{Evaluating the limit}} \cr
& 1000\mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{{0.09}}{n}} \right)^n} = {\left( {1 + \frac{{0.09}}{\infty }} \right)^\infty } = {1^\infty } \cr
& {\text{This limit has the form }}{1^\infty }{\text{ }} \cr
& {\left( {1 + \frac{{0.09}}{n}} \right)^n} = {e^{\ln {{\left( {1 + \frac{{0.09}}{n}} \right)}^n}}},{\text{ then}} \cr
& \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{{0.09}}{n}} \right)^n} = \mathop {\lim }\limits_{n \to \infty } {e^{\ln {{\left( {1 + \frac{{0.09}}{n}} \right)}^n}}} = {e^{\mathop {\lim }\limits_{n \to \infty } n\ln \left( {1 + \frac{{0.09}}{n}} \right)}} \cr
& {\text{The first step is to evaluate }} \cr
& L = \mathop {\lim }\limits_{n \to \infty } n\left( {1 + \frac{{0.09}}{n}} \right) = \mathop {\lim }\limits_{n \to \infty } \frac{{\ln \left( {1 + \frac{{0.09}}{n}} \right)}}{{1/n}} = \frac{{\ln \left( {1 + \frac{{0.09}}{\infty }} \right)}}{{1/\infty }} = \frac{0}{0} \cr
& {\text{Using L'Hopital's rule}} \cr
& L = \mathop {\lim }\limits_{n \to \infty } \frac{{\frac{d}{{dn}}\left[ {\ln \left( {1 + \frac{{0.09}}{n}} \right)} \right]}}{{\frac{d}{{dn}}\left[ {1/n} \right]}} = \mathop {\lim }\limits_{n \to \infty } \frac{{\frac{{ - 0.09/{n^2}}}{{1 + 0.09/n}}}}{{\left( { - 1/{n^2}} \right)}} \cr
& = \mathop {\lim }\limits_{n \to \infty } \frac{{0.09}}{{1 + 0.09/n}} \cr
& = \frac{{0.09}}{{1 + 0.09/\infty }} = 0.09 \cr
& {\text{Therefore,}} \cr
& \mathop {\lim }\limits_{n \to \infty } {e^{n\ln \left( {1 + \frac{{0.09}}{n}} \right)}} = {e^{\mathop {\lim }\limits_{n \to \infty } n\ln \left( {1 + \frac{{0.09}}{n}} \right)}} = {e^{0.09}} \cr
& 1000\mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{{0.09}}{n}} \right)^n} = 1000{e^{0.09}} \approx 1094.1742 \cr} $$
