# 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.

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You have to rationalize the denominator? Says who? –  anon Sep 6 '11 at 21:48
Yes, multiply and divide by the conjugate. You won't have to rationalize the denominator. Try it. –  lhf Sep 6 '11 at 21:49
I thought it was a rule. –  user138246 Sep 6 '11 at 21:50
I multiplied by the conjugate and I think I got x on the top since the rest cancelled out and then conjugate on the bottom. I think that reduces to 3 if I divide everything by x, which gives me the wrong answer. –  user138246 Sep 6 '11 at 21:55
Possible Duplicate: math.stackexchange.com/questions/30040/… –  JavaMan Sep 6 '11 at 22:26

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

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I don't understand how 9x^2 turns into (9+1/x) –  user138246 Sep 6 '11 at 21:59
@Jordan You divide the numerator and denominator by $x$. The first term alone looks like $\frac{1}{x}\sqrt{9x^2+x}$. Can you then bring the $1/x$ inside the square root? What will happen to it then? –  Srivatsan Sep 6 '11 at 22:02
@Jordan: I edited it with an explanation. –  anon Sep 6 '11 at 22:02
I don't understand where the 1/x infront came from, would that make the whole numerator 0? –  user138246 Sep 6 '11 at 22:06
@Jordan: If you divide $\sqrt{9x^2+x}$ by $x$ you get $\frac{1}{x}\sqrt{9x^2+x}$. What makes you think this makes the numerator $0$? –  anon Sep 6 '11 at 22:07

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}

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What is that O? –  user138246 Sep 7 '11 at 1:58
@Jordan: that is the "big-$O$". Here, $O\left(\frac{1}{x^2}\right)$ represents a quantity for which there is a $C>0$ so that the absolute value of that quantity is smaller than $C\times\frac{1}{x^2}$ as $x\to\infty$. –  robjohn Sep 7 '11 at 2:13

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:

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

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