Finding $\lim\limits_{x \to \infty} (\sqrt{9x^2+x} - 3x)$

I am trying to find the limit and I know that I have to do something to get a denominator but I can't make it work. There is a specific thing I have to do to this problem to make it work but I am not sure what that is. I know what I am doing is not wrong mathematically, it just doesn't help me get the answer.

$$\lim_{x \to \infty} (\sqrt{9x^2+x} - 3x)$$

Not sure where to go with this, multiply it all by the conjugate? I tried that but I know that won't help because I have to rationalize the denominator. I know I can't multiply by anything really until I get rid of the sqrt and I don't know how.

$$\lim_{x\to\infty}(\sqrt{9x^2+x}-3x)$$ $$=\lim_{x\to\infty}\frac{x}{\sqrt{9x^2+x}+3x}.$$ Divide numerator and denominator by $x$, $$=\lim_{x\to\infty}\frac{1}{\frac{1}{x}\sqrt{9x^2+x}+3}$$ note $\frac{1}{x}\sqrt{9x^2+x}=\sqrt{\frac{9x^2+x}{x^2}}=\sqrt{9+\frac{1}{x}}$ so this reduces to $$=\lim_{x\to\infty}\frac{1}{\sqrt{9+\frac{1}{x}}+3}$$ $$=\frac{1}{\sqrt{9}+3}=\frac{1}{6}.$$ And just FYI this ground was (more or less) covered in this question.

Factoring out the $9x^2$ from inside the radical, then applying the binomial theorem (or Taylor series or whatever) to get the partial series: $\sqrt{1+x}=1+\frac{x}{2}+O(x^2)$, we get $$ \begin{align} \lim_{x\to\infty}(\sqrt{9x^2+x}-3x)&=\lim_{x\to\infty}3x\left(\sqrt{1+\frac{1}{9x}}-1\right)\\ &=\lim_{x\to\infty}3x\left(\frac{1}{18x}+O\left(\frac{1}{x^2}\right)\right)\\ &=\frac{1}{6} \end{align} $$

I'll use your previous question: Finding $\lim\limits_{h \to 0}\frac{(\sqrt{9+h} -3)}{h}$ , where you got the limit was $1/6$.

Indeed, to compute $\lim\limits_{x\to +\infty} (\sqrt{9x^2+x}-3x)$, if you let $x=1/h$, then as $x\to +\infty$ you get $h\to 0^+$. So:

$$\lim_{x\to +\infty} (\sqrt{9x^2+x}-3x)= \lim_{h\to 0^+} \left(\sqrt{\frac{9}{h^2}+\frac{1}{h}}-\frac{3}{h}\right) = \lim_{h\to 0^+} \left(\sqrt{\frac{1}{h^2}(9+h)}-\frac{3}{h}\right) = \lim_{h\to 0^+} \frac{\sqrt{9+h}-3}{h}= \frac{1}{6}$$