Finding the limit of $\sqrt{9x^2+x} -3x$ at infinity To find the limit
$$\lim_{x\to\infty} (\sqrt{9x^2+x} -3x)$$
Basically I simplified this down to
$$\lim_{x\to\infty} \frac{1}{\sqrt{9+1/x}+3x}$$
And I am unaware of what to do next. I tried to just sub in infinity and I get an answer of $0$ , since $1 / \infty = 0$. However, on symbolab, when I enter the problem it gives me an answer of $1/6$.
Can anyone please explain to me what I need to do from this point? I haven't learned l'Hopitals rule yet so please don't suggest that. Thanks

For $x>0$ we have 
$$\sqrt{9x^2+x}-3x=\frac{\big(\sqrt{9x^2+x}-3x\big)\big(\sqrt{9x^2+x}+3x\big)}{\sqrt{9x^2+x}+3x}= \\ =\frac{9x^2+x-9x^2}{\sqrt{9x^2+x}+3x}=\frac{x}{\sqrt{9x^2+x}+3x}=\frac{1}{\sqrt{9+\frac{1}{x}}+3}=$$
Thus:
$$\lim_{x\rightarrow\infty}\big(\sqrt{9x^2+x}-3x\big)=\lim_{x\rightarrow\infty}\frac{1}{\sqrt{9+\frac{1}{x}}+3}=\frac{1}{6}$$
