Answer
$$-\frac{1}{6}$$
Work Step by Step
\begin{aligned} \lim _{x \rightarrow \infty}(\sqrt{9 x^{2}-x}-3 x) &=\lim _{x \rightarrow \infty}[\sqrt{9 x^{2}-x}-3 x] \cdot\left[\frac{\sqrt{9 x^{2}-x}+3 x}{\sqrt{9 x^{2}-x}+3 x}\right]\\
&=\lim _{x \rightarrow \infty} \frac{\left(9 x^{2}-x\right)-\left(9 x^{2}\right)}{\sqrt{9 x^{2}-x+3 x}}\\
&=\lim _{x \rightarrow \infty} \frac{-x}{\sqrt{9 x^{2}-x}+3 x} \\
&=\lim _{x \rightarrow \infty} \frac{\frac{-x}{x}}{\sqrt{\frac{9 x^{2}-x}{x^{2}}+\frac{3 x}{x}}}\\
&=\lim _{x \rightarrow \infty} \frac{-1}{\sqrt{9-\frac{1}{x}}+3}\\
&=\frac{-1}{3+3}\\
&=-\frac{1}{6}
\end{aligned}
