anonymous
  • anonymous
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Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\lim_{x \rightarrow -\infty}\frac{ x }{ \sqrt{9x^{2}+x}+3x }\]
anonymous
  • anonymous
The conjugate turns it into: \[\lim_{x \rightarrow -\infty}(\sqrt{9x^{2}+x}-3x)\] I don't see how that helps. And that is incorrect @zzr0ck3r The answer is \(\infty\)
ganeshie8
  • ganeshie8
rationalize the denominator and simply evaluate the limit ?

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zzr0ck3r
  • zzr0ck3r
hmm \[\lim_{x \rightarrow -\infty}\frac{ x }{ \sqrt{9x^{2}+x}+3x }\\=\lim_{x \rightarrow -\infty}\dfrac{\frac{1}{x}}{\frac{1}{x}}\frac{ x }{ \sqrt{9x^{2}+x}+3x }\\=\lim_{x\rightarrow -\infty}\dfrac{1}{\sqrt{9+\frac{1}{x}}+3}=\dfrac{1}{6}\]
IrishBoy123
  • IrishBoy123
well the square root on the bottom gives 2 answers .... ???? \(\pm 3 + 3\)
anonymous
  • anonymous
Yes, 1/6 works as the limit to positive infinity, yet not to minus infinity. It's weird.
zzr0ck3r
  • zzr0ck3r
You sure its \(\infty\)?
IrishBoy123
  • IrishBoy123
plot it!
ganeshie8
  • ganeshie8
https://www.desmos.com/calculator/g0clqqhnx8
zzr0ck3r
  • zzr0ck3r
The inside of the square root would be positive for big x
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=lim%28x%2F%28sqrt%289x^2%2Bx%29%2B3x%29%29+x+to+-infinity
zzr0ck3r
  • zzr0ck3r
woops, got side tracked by https://www.facebook.com/JustEatingRealFood/videos/711816552279115/?fref=nf
anonymous
  • anonymous
Well, this is what I assumed would have been the process: \[\lim_{x \rightarrow -\infty}\frac{ x }{ \sqrt{9x^{2}+x}+3x } = \lim_{x \rightarrow -\infty}\frac{ x }{ |x|\sqrt{9 +\frac{ 1 }{ x }}+3x}\] \(\sqrt{x^{2}} = |x|\), but since x is going to minus infinity, I thought the absolute value would become -x, in which I could continue from there. But that gives me \(\frac{1}{-3+3}\), which is blah! Haha
anonymous
  • anonymous
Happy dance :D
zzr0ck3r
  • zzr0ck3r
I hate calculus lol
zzr0ck3r
  • zzr0ck3r
@ganeshie8 will save the day
ganeshie8
  • ganeshie8
\(\lim\limits_{x \rightarrow -\infty}(\sqrt{9x^{2}+x}-3x)\) Ahh right, absolute bars seems to be the key \[\sqrt{9x^2+x} =\sqrt{\left| 3x+\frac{1}{6}\right|^2 -\frac{1}{36}} \] as \(x\to -\infty\), above expression approaches \(\left|3x+\frac{1}{6}\right| = -3x-\frac{1}{6}\)
anonymous
  • anonymous
Okay, so I see how the square root approaches that. Do we still have the \(3x\) term? Not that it would matter, but making sure whether or not we only considered the root or the entire expression.
ganeshie8
  • ganeshie8
\(\lim\limits_{x \rightarrow -\infty}(\sqrt{9x^{2}+x}-3x) = \lim\limits_{x \rightarrow -\infty}(-3x-\frac{1}{6}-3x)=\infty\) \(\lim\limits_{x \rightarrow \infty}(\sqrt{9x^{2}+x}-3x) = \lim\limits_{x \rightarrow \infty}(3x+\frac{1}{6}-3x)=\frac{1}{6}\)
anonymous
  • anonymous
Very clever @ganeshie8 :) I suppose I would just have to make the same argument you did. And you showed me a new trick to consider with the absolute values :D \(x^{2} \rightarrow\ |x|^{2}\)
anonymous
  • anonymous
Thanks again, like many many times before, lol.
ganeshie8
  • ganeshie8
np:)

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