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adam32885

  • one year ago

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

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  1. kirbykirby
    • one year ago
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    Do you know L'Hopital's rule yet?

  2. adam32885
    • one year ago
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    no not yet

  3. kirbykirby
    • one year ago
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    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
    • one year ago
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    I know i need to factor the top and bottom but i am drawing a blank on factoring

  5. adam32885
    • one year ago
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    oh duh so it would t-1?

  6. kirbykirby
    • one year ago
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    oh wait that doesn't help sorry. I'm so used of thinking of limits going to infinity

  7. kirbykirby
    • one year ago
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    Ok Yes I suppose factoring is another way

  8. adam32885
    • one year ago
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    that is where we are currently at is trying to factor them out

  9. kirbykirby
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    ok gotcha thank you

  12. kirbykirby
    • one year ago
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    Let me know if you are still stuck after :)

  13. adam32885
    • one year ago
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    i will thanks again

  14. kirbykirby
    • one year ago
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    np

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