Answer
\[2\]
Work Step by Step
\[\begin{gathered}
f\left( x \right) = \frac{{4x + 3}}{{2x - 1}} \hfill \\
{\text{Evaluate }}f\left( x \right){\text{ for the given values and complete the table}}{\text{.}} \hfill \\
x = {10^0} \to f\left( {{{10}^0}} \right) = \frac{{4\left( {{{10}^0}} \right) + 3}}{{2\left( {{{10}^0}} \right) - 1}} = 7 \hfill \\
x = {10^1} \to f\left( {{{10}^1}} \right) = \frac{{4\left( {{{10}^1}} \right) + 3}}{{2\left( {{{10}^1}} \right) - 1}} \approx 2.263 \hfill \\
x = {10^2} \to f\left( {{{10}^2}} \right) = \frac{{4\left( {{{10}^2}} \right) + 3}}{{2\left( {{{10}^2}} \right) - 1}} \approx 2.2025 \hfill \\
x = {10^3} \to f\left( {{{10}^3}} \right) = \frac{{4\left( {{{10}^3}} \right) + 3}}{{2\left( {{{10}^3}} \right) - 1}} \approx 2.0025 \hfill \\
x = {10^4} \to f\left( {{{10}^4}} \right) = \frac{{4\left( {{{10}^4}} \right) + 3}}{{2\left( {{{10}^4}} \right) - 1}} = 2.0003 \hfill \\
x = {10^5} \to f\left( {{{10}^5}} \right) = \frac{{4\left( {{{10}^5}} \right) + 3}}{{2\left( {{{10}^5}} \right) - 1}} = 2 \hfill \\
x = {10^6} \to f\left( {{{10}^6}} \right) = \frac{{4\left( {{{10}^6}} \right) + 3}}{{2\left( {{{10}^6}} \right) - 1}} = 2 \hfill \\
\boxed{\begin{array}{*{20}{c}}
x&{{{10}^0}}&{{{10}^1}}&{{{10}^2}}&{{{10}^3}}&{{{10}^4}}&{{{10}^5}}&{{{10}^6}} \\
{f\left( x \right)}&7&{2.263}&{2.2025}&{2.0025}&{2.0003}&2&2
\end{array}} \hfill \\
{\text{Therefore,}} \hfill \\
\mathop {\lim }\limits_{x \to \infty } f\left( x \right) = \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{4x + 3}}{{2x - 1}}} \right) = 2 \hfill \\
{\text{Graph}} \hfill \\
\end{gathered} \]
