A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

Limit Question

  • This Question is Closed
  1. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  2. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    rationalize the denominator and simply evaluate the limit ?

  4. zzr0ck3r
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  6. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  7. zzr0ck3r
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    You sure its \(\infty\)?

  8. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    plot it!

  9. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  10. zzr0ck3r
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  11. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  12. zzr0ck3r
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  13. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Happy dance :D

  15. zzr0ck3r
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I hate calculus lol

  16. zzr0ck3r
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    @ganeshie8 will save the day

  17. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 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 \(\left|3x+\frac{1}{6}\right| = -3x-\frac{1}{6}\)

  18. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

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

  20. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Thanks again, like many many times before, lol.

  22. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    np:)

  23. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.