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anonymous
 one year ago
Limit Question
anonymous
 one year ago
Limit Question

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\lim_{x \rightarrow \infty}\frac{ x }{ \sqrt{9x^{2}+x}+3x }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The 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
 one year ago
Best ResponseYou've already chosen the best response.3rationalize the denominator and simply evaluate the limit ?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1hmm \[\lim_{x \rightarrow \infty}\frac{ x }{ \sqrt{9x^{2}+x}+3x }\\=\lim_{x \rightarrow \infty}\dfrac{\frac{1}{x}}{\frac{1}{x}}\frac{ x }{ \sqrt{9x^{2}+x}+3x }\\=\lim_{x\rightarrow \infty}\dfrac{1}{\sqrt{9+\frac{1}{x}}+3}=\dfrac{1}{6}\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1well the square root on the bottom gives 2 answers .... ???? \(\pm 3 + 3\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, 1/6 works as the limit to positive infinity, yet not to minus infinity. It's weird.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1You sure its \(\infty\)?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1The inside of the square root would be positive for big x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=lim%28x%2F%28sqrt%289x^2%2Bx%29%2B3x%29%29+x+to+infinity

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1woops, got side tracked by https://www.facebook.com/JustEatingRealFood/videos/711816552279115/?fref=nf

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, 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

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1@ganeshie8 will save the day

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3\(\lim\limits_{x \rightarrow \infty}(\sqrt{9x^{2}+x}3x)\) 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 \(\left3x+\frac{1}{6}\right = 3x\frac{1}{6}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, 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
 one year ago
Best ResponseYou've already chosen the best response.3\(\lim\limits_{x \rightarrow \infty}(\sqrt{9x^{2}+x}3x) = \lim\limits_{x \rightarrow \infty}(3x\frac{1}{6}3x)=\infty\) \(\lim\limits_{x \rightarrow \infty}(\sqrt{9x^{2}+x}3x) = \lim\limits_{x \rightarrow \infty}(3x+\frac{1}{6}3x)=\frac{1}{6}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Very 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
 one year ago
Best ResponseYou've already chosen the best response.0Thanks again, like many many times before, lol.
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