A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

owlet

  • one year ago

Question below:

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

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

    what's the next step?

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

    @SolomonZelman @Loser66 @freckles

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

    \(\Large\color{black}{\displaystyle\lim_{x \to~ -\infty}\frac{\sqrt{2x^2+x+1}}{3x-1}}\)

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

    you can do this: \(\Large\color{black}{\displaystyle\lim_{x \to~ -\infty}\frac{\sqrt{2x^2+x+1\color{white}{\LARGE |}}}{3x-1}}\) \(\Large\color{black}{\displaystyle\lim_{x \to~ -\infty}\frac{\sqrt{2x^2+x+1\color{white}{\LARGE |}}}{\sqrt{(3x-1)^2}}}\) \(\Large\color{black}{\displaystyle\lim_{x \to~ -\infty}\sqrt{\frac{2x^2+x+1}{(3x-1)^2}}}\)

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

    \(\Large\color{black}{\displaystyle\lim_{x \to~ -\infty}\sqrt{\frac{2x^2+x+1}{9x^2-6x+1}}=\sqrt{2/9}=\sqrt{2}/3}\)

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

    you can bring the limit inside the root, and do L'Hospital's rule twice.

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

    how will i apply x approaches -infinity though?

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

    Ok, \(\large\color{black}{\displaystyle\lim_{x \to~ -\infty}\sqrt{\frac{2x^2+x+1}{9x^2-6x+1}}=\sqrt{\lim_{x \to~ -\infty}\frac{2x^2+x+1}{9x^2-6x+1}}}\)

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

    \(\large\color{black}{\displaystyle \sqrt{\lim_{x \to~ -\infty}\frac{2x^2+x+1}{9x^2-6x+1}}}\) You could apply L'Hospital's rule at this point, because as x\(\to -\infty\), the top and bottom tend to ∞. (an ∞/∞ case)

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

    differentiate on top and bottom and you get \(\large\color{black}{\displaystyle \sqrt{\lim_{x \to~ -\infty}\frac{4x+1}{18x-6}}}\)

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

    but we haven't learned l'hospital's rule yet

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

    Oh, so you aren't allowed to use it? (But do you know that rule, though?)

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

    no, we're not allowed and I don't know that rule also

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

    \(\large\color{black}{\displaystyle \sqrt{\lim_{x \to~ -\infty}\frac{2x^2+x+1}{9x^2-6x+1}}}\) You can just conclude the answer based on the leading coefficients, then. (The limit will be equal to 2/9 and the square root of 2/9 is going to be equal to √2 /3

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

    It really did seem like an L'Hospital's rule kind of a problem, I apolgoize for mentioning it.

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

    sorry i was off. I just finished my class. will it always work though? for those kind of problems? Just get the leading coefficients? I want to learn that rule.. where can I learn it? maybe I can just search it up on the internet.

  18. 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.