Answer
\[ - 3\]
Work Step by Step
\[\begin{gathered}
f\left( x \right) = \frac{{ - 6x}}{{\sqrt {4{x^2} + 5} }} \hfill \\
{\text{Evaluate }}f\left( x \right){\text{ for the given values and complete the table}}{\text{.}} \hfill \\
x = {10^0} \to f\left( {{{10}^0}} \right) = \frac{{ - 6\left( {{{10}^0}} \right)}}{{\sqrt {4{{\left( {{{10}^0}} \right)}^2} + 5} }} = - 2 \hfill \\
x = {10^1} \to f\left( {{{10}^1}} \right) = \frac{{ - 6\left( {{{10}^1}} \right)}}{{\sqrt {4{{\left( {{{10}^1}} \right)}^2} + 5} }} \approx - 2.981 \hfill \\
x = {10^2} \to f\left( {{{10}^2}} \right) = \frac{{ - 6\left( {{{10}^2}} \right)}}{{\sqrt {4{{\left( {{{10}^2}} \right)}^2} + 5} }} \approx - 2.9998 \hfill \\
x = {10^3} \to f\left( {{{10}^3}} \right) = \frac{{ - 6\left( {{{10}^3}} \right)}}{{\sqrt {4{{\left( {{{10}^3}} \right)}^2} + 5} }} \approx - 2.999998 \hfill \\
x = {10^4} \to f\left( {{{10}^4}} \right) = \frac{{ - 6\left( {{{10}^4}} \right)}}{{\sqrt {4{{\left( {{{10}^4}} \right)}^2} + 5} }} \approx - 3 \hfill \\
x = {10^5} \to f\left( {{{10}^5}} \right) = \frac{{ - 6\left( {{{10}^5}} \right)}}{{\sqrt {4{{\left( {{{10}^5}} \right)}^2} + 5} }} \approx - 3 \hfill \\
x = {10^6} \to f\left( {{{10}^6}} \right) = \frac{{ - 6\left( {{{10}^6}} \right)}}{{\sqrt {4{{\left( {{{10}^6}} \right)}^2} + 5} }} \approx - 3 \hfill \\
\boxed{\begin{array}{*{20}{c}}
x&{f\left( x \right)} \\
{{{10}^0}}&{ - 2} \\
{{{10}^1}}&{ - 2.981} \\
{{{10}^2}}&{ - 2.9998} \\
{{{10}^3}}&{ - 2.999998} \\
{{{10}^4}}&{ - 3} \\
{{{10}^5}}&{ - 3} \\
{{{10}^6}}&{ - 3}
\end{array}} \hfill \\
{\text{Therefore,}} \hfill \\
\mathop {\lim }\limits_{x \to \infty } f\left( x \right) = \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{ - 6x}}{{\sqrt {4{x^2} + 5} }}} \right) = - 3 \hfill \\
{\text{Graph}} \hfill \\
\end{gathered} \]
