Answer
\[5\]
Work Step by Step
\[\begin{gathered}
f\left( x \right) = 5 - \frac{1}{{{x^2} + 1}} \hfill \\
{\text{Evaluate }}f\left( x \right){\text{ for the given values and complete the table}}{\text{.}} \hfill \\
x = {10^0} \to f\left( {{{10}^0}} \right) = 5 - \frac{1}{{{{\left( {{{10}^0}} \right)}^2} + 1}} = 4.5 \hfill \\
x = {10^1} \to f\left( {{{10}^1}} \right) = 5 - \frac{1}{{{{\left( {{{10}^1}} \right)}^2} + 1}} \approx 4.999 \hfill \\
x = {10^2} \to f\left( {{{10}^2}} \right) = 5 - \frac{1}{{{{\left( {{{10}^2}} \right)}^2} + 1}} \approx 4.9999 \hfill \\
x = {10^3} \to f\left( {{{10}^3}} \right) = 5 - \frac{1}{{{{\left( {{{10}^3}} \right)}^2} + 1}} \approx 4.999999 \hfill \\
x = {10^4} \to f\left( {{{10}^4}} \right) = 5 - \frac{1}{{{{\left( {{{10}^4}} \right)}^2} + 1}} \approx 5 \hfill \\
x = {10^5} \to f\left( {{{10}^5}} \right) = 5 - \frac{1}{{{{\left( {{{10}^5}} \right)}^2} + 1}} \approx 5 \hfill \\
x = {10^6} \to f\left( {{{10}^6}} \right) = 5 - \frac{1}{{{{\left( {{{10}^6}} \right)}^2} + 1}} \approx 5 \hfill \\
\boxed{\begin{array}{*{20}{c}}
x&{f\left( x \right)} \\
{{{10}^0}}&{4.5} \\
{{{10}^1}}&{4.999} \\
{{{10}^2}}&{4.9999} \\
{{{10}^3}}&{4.999999} \\
{{{10}^4}}&5 \\
{{{10}^5}}&5 \\
{{{10}^6}}&5
\end{array}} \hfill \\
{\text{Therefore,}} \hfill \\
\mathop {\lim }\limits_{x \to \infty } f\left( x \right) = \mathop {\lim }\limits_{x \to \infty } \left( {5 - \frac{1}{{{x^2} + 1}}} \right) = 5 \hfill \\
{\text{Graph}} \hfill \\
\end{gathered} \]
