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