Finding the limit of $\sqrt{9x^2+x} -3x$ at infinity

Question

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

The x that you highlighted in your simplification is not supposed to be there. — Kaynex, Feb 19 '16 at 00:47
You might have a look at a more general question here and also at other post which are linked there. In particular the two questions which were suggested as duplicates so far seem rather similar to your question: Finding $\lim\limits_{x \to \infty} \sqrt{9x^2+x} - 3x$ and Limit of $\sqrt{4x^2 + 3x} - 2x$ as $x \to \infty$ — Martin Sleziak, Feb 19 '16 at 02:38

score 3 · Answer 1 · answered Feb 19 '16 at 01:04

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: $$\lim_{x\rightarrow\infty}\big(\sqrt{9x^2+x}-3x\big)=\lim_{x\rightarrow\infty}\frac{1}{\sqrt{9+\frac{1}{x}}+3}=\frac{1}{6}$$

Finding the limit of $\sqrt{9x^2+x} -3x$ at infinity

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