Question

Limit Question

Answer 1

\[\lim_{x \rightarrow -\infty}\frac{ x }{ \sqrt{9x^{2}+x}+3x }\]

Answer 2

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 \(\infty\)

Answer 3

rationalize the denominator and simply evaluate the limit ?

Answer 4

hmm

\[\lim_{x \rightarrow -\infty}\frac{ x }{ \sqrt{9x^{2}+x}+3x }\\=\lim_{x \rightarrow -\infty}\dfrac{\frac{1}{x}}{\frac{1}{x}}\frac{ x }{ \sqrt{9x^{2}+x}+3x }\\=\lim_{x\rightarrow -\infty}\dfrac{1}{\sqrt{9+\frac{1}{x}}+3}=\dfrac{1}{6}\]

Answer 5

well the square root on the bottom gives 2 answers .... ????

\(\pm 3 + 3\)

Answer 6

Yes, 1/6 works as the limit to positive infinity, yet not to minus infinity. It's weird.

Answer 7

You sure its \(\infty\)?

Answer 8

plot it!

Answer 9

https://www.desmos.com/calculator/g0clqqhnx8

Answer 10

The inside of the square root would be positive for big x

Answer 11

http://www.wolframalpha.com/input/?i=lim%28x%2F%28sqrt%289x^2%2Bx%29%2B3x%29%29+x+to+-infinity

Answer 12

woops, got side tracked by

https://www.facebook.com/JustEatingRealFood/videos/711816552279115/?fref=nf

Answer 13

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

Answer 14

Happy dance :D

Answer 15

I hate calculus lol

Answer 16

@ganeshie8 will save the day

Answer 17

\(\lim\limits_{x \rightarrow -\infty}(\sqrt{9x^{2}+x}-3x)\)

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 \(\left|3x+\frac{1}{6}\right| = -3x-\frac{1}{6}\)

Answer 18

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.

Answer 19

\(\lim\limits_{x \rightarrow -\infty}(\sqrt{9x^{2}+x}-3x) = \lim\limits_{x \rightarrow -\infty}(-3x-\frac{1}{6}-3x)=\infty\)


\(\lim\limits_{x \rightarrow \infty}(\sqrt{9x^{2}+x}-3x) = \lim\limits_{x \rightarrow \infty}(3x+\frac{1}{6}-3x)=\frac{1}{6}\)

Answer 20

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 \(x^{2} \rightarrow\ |x|^{2}\)

Answer 21

Thanks again, like many many times before, lol.

Answer 22

np:)