A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

a question about limit.

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

    \[\lim_{x \rightarrow -\infty} (2x-1)\left[ \frac{ -5 }{ x-2 } \right]\]

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

    @ganeshie8 .

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

    is that a floor function or just being multiplied

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

    multiplied

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

    Oh why not just multiply through and then expand \[\frac{ -5(2x-1) }{ x-2 } = \frac{ -10x }{ x-2 }+\frac{ 5 }{ x-2 }\] then you can take the limits separately I was being lazy by not putting limits you shouldn't do that

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

    The first one -10x/(x-2) might have some trouble but you can let \[x-2 = x\left( 1-\frac{ 2 }{ x } \right)\] then you can cancel out the x's and then just plug in stuff with limits

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

    no .-5/(x-2) is in brackets .that means [3.14]=3

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

    What?

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

    So that is a floor function..

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

    Eh I have to go but with this problem @jtvatsim and @ganeshie8 can help you, they're great.

  11. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    Notice that the function \(\left[\dfrac{-5}{x-2}\right] \) is identically equal to \(0\) for \(x\lt -3\). Since this is just the definition of floor function, and our interest is in finding limit as \(x\to-\infty\), we may replace that piece by \(0\) : \[\lim_{x \rightarrow -\infty} (2x-1)\left[ \frac{ -5 }{ x-2 } \right] = \lim_{x \rightarrow -\infty} (2x-1)*0 = 0\]

  12. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    Incase if you want to find the limit \(x\to\infty\), similar reasoning yields : \[\lim_{x \rightarrow -\infty} (2x-1)\left[ \frac{ -5 }{ x-2 } \right] = \lim_{x \rightarrow \infty} (2x-1)*(-1) = -\infty\]

  13. 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.