## anonymous one year ago Limit Question

1. anonymous

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

2. 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 $$\infty$$

3. ganeshie8

rationalize the denominator and simply evaluate the limit ?

4. zzr0ck3r

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

5. IrishBoy123

well the square root on the bottom gives 2 answers .... ???? $$\pm 3 + 3$$

6. anonymous

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

7. zzr0ck3r

You sure its $$\infty$$?

8. IrishBoy123

plot it!

9. ganeshie8
10. zzr0ck3r

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

11. anonymous
12. zzr0ck3r

woops, got side tracked by https://www.facebook.com/JustEatingRealFood/videos/711816552279115/?fref=nf

13. 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

14. anonymous

Happy dance :D

15. zzr0ck3r

I hate calculus lol

16. zzr0ck3r

@ganeshie8 will save the day

17. 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 $$\left|3x+\frac{1}{6}\right| = -3x-\frac{1}{6}$$

18. 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.

19. ganeshie8

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

20. 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 $$x^{2} \rightarrow\ |x|^{2}$$

21. anonymous

Thanks again, like many many times before, lol.

22. ganeshie8

np:)