1. **State the problem:** Find the limit as $x$ approaches infinity of the expression $$\sqrt{9x^2 + x} - 3x.$$\n\n2. **Recall the formula and rules:** When dealing with limits involving square roots and large $x$, it helps to rationalize the expression or factor out the highest power of $x$ inside the root to simplify.\n\n3. **Rewrite the expression:** $$\sqrt{9x^2 + x} - 3x = \frac{(\sqrt{9x^2 + x} - 3x)(\sqrt{9x^2 + x} + 3x)}{\sqrt{9x^2 + x} + 3x} = \frac{9x^2 + x - 9x^2}{\sqrt{9x^2 + x} + 3x} = \frac{x}{\sqrt{9x^2 + x} + 3x}.$$\n\n4. **Simplify the denominator:** Factor $x$ inside the square root: $$\sqrt{9x^2 + x} = \sqrt{x^2(9 + \frac{1}{x})} = x\sqrt{9 + \frac{1}{x}}.$$\n\n5. **Substitute back:** $$\frac{x}{x\sqrt{9 + \frac{1}{x}} + 3x} = \frac{x}{x(\sqrt{9 + \frac{1}{x}} + 3)} = \frac{1}{\sqrt{9 + \frac{1}{x}} + 3}.$$\n\n6. **Evaluate the limit as $x \to \infty$:** Since $\frac{1}{x} \to 0$, we get $$\lim_{x \to \infty} \frac{1}{\sqrt{9 + 0} + 3} = \frac{1}{3 + 3} = \frac{1}{6}.$$\n\n**Final answer:** $$\boxed{\frac{1}{6}}.$$