Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

adam32885

  • 3 years ago

lim as t approaches 1 ((t^4-1)/(t^3-1)

  • This Question is Closed
  1. kirbykirby
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Do you know L'Hopital's rule yet?

  2. adam32885
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    no not yet

  3. kirbykirby
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Ok Well you can proceed this way: You want to try and get rid of as many powers as you can. Why not try dividing the numerator and denominator by t^3?

  4. adam32885
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I know i need to factor the top and bottom but i am drawing a blank on factoring

  5. adam32885
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh duh so it would t-1?

  6. kirbykirby
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    oh wait that doesn't help sorry. I'm so used of thinking of limits going to infinity

  7. kirbykirby
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Ok Yes I suppose factoring is another way

  8. adam32885
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    that is where we are currently at is trying to factor them out

  9. kirbykirby
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Ok t^4-1 is easy to factor. If you think of letting t^2 = x, then you get (x^2-1) which is easily factorable.

  10. kirbykirby
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    t^3-1 is a bit trickier. Do you remember that one way to this is to try finding all factors of your constant term (here it's -1) and plugging it into t^3-1 and see which ones give you zero. This number (say "a") will then be a factor of the form (t-a). Then, you can use long polynomial division to find out the other factor by dividing t^3-1 with (t-a)

  11. adam32885
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok gotcha thank you

  12. kirbykirby
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Let me know if you are still stuck after :)

  13. adam32885
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i will thanks again

  14. kirbykirby
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    np

  15. 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