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Limit (L'hopitals)

  1. Mar 2, 2013 #1
    1. The problem statement, all variables and given/known data
    evaluate the limit using l'hopitals rule.


    2. Relevant equations
    limit x->9+ (1/(x-9) - 1/ln(x-8))


    3. The attempt at a solution
    I tried taking the derivative of the numerator and denominator in each, getting 0 as an answer, but it wasn't correct.
     
  2. jcsd
  3. Mar 2, 2013 #2

    CAF123

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    Reexpress that as a single fraction then |'Hopital's Rule becomes valid.
     
  4. Mar 3, 2013 #3
    ln(x-8) - x - 9
    ------------------ 0/9 ---> should be zero right?
    x - 9ln(x-8)


    anyways, apply l'hopitals (take derivative of numerator and denominator):


    (1/x-8) - 1
    ------------ = 0/-8 = 0
    1 - 9(1/x-8)


    answer should be, but i can't figure out how: (-1/2)
     
  5. Mar 3, 2013 #4

    eumyang

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    Be careful with your algebra. I'm getting
    [tex]\frac{\ln (x-8) - x + 9}{(x - 9)\ln(x - 8)}[/tex]
    Of course, as x approaches 9 from the right the fraction goes to 0/0.
     
  6. Mar 3, 2013 #5

    Dick

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    I really wish you would write things using more parentheses where needed. You have (log(x-8)-(x-9))/((x-9)*log(x-8)). The denominator is a product. Use the product rule. You'll should get another 0/0 limit. That just means you should use l'Hopital again.
     
  7. Mar 3, 2013 #6
    now i'm getting 1 as the answer, after correcting my algebra, and applying l'hopitals again.

    (1/(-x+8)^2)
    lim(x->9+ ------------------
    (x-9)(1/(-x+8)^2) + (1/(x-8)) - 9ln(x-8)



    = (1/1)/1
     
  8. Mar 3, 2013 #7

    eumyang

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    You should also show all steps in your work, not just the last one. It's harder to detect where you went wrong. You should also learn LaTeX -- it's very hard to read what you wrote above.

    What did you get when you took the first derivative? Before taking the second derivative, you should simplify the expression by multiplying top and bottom by (x - 8). That way you won't have to deal with complex fractions.
     
  9. Mar 3, 2013 #8
    sorry i cna't figure the latex out.


    1. limit x->9+ (1/(x-9) - 1/ln(x-8))

    2. ln(x-8) - x + 9
    ----------------- Common denominator. (0/0)
    (x-9) * (ln(x-8))



    3. (1/(x-8)) - 1
    ----------------- derivatives of numerator and denominator. using product rule in denominator. (0/0)
    (x-9) * (1/(x-8)) + 1(1/ln(x-8))



    4. (-1/(x-8)^2)
    --------------------- derivatives of numerator and denominator. using product rule in denominator. (1/1)
    (x-9)* (-1/(x-8)^2) + (1/(x-8) + (-1/(x-8)*ln(x-8)^2)
     
    Last edited: Mar 3, 2013
  10. Mar 3, 2013 #9

    eumyang

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    If I can do it, so can you!


    This is wrong.

    After you fix step #3, as I said previously, before you take the derivative again, multiply the numerator and denominator by (x-8), so that you won't have to deal with complex fractions.
     
  11. Mar 3, 2013 #10
    what's wrong with it?
     
  12. Mar 3, 2013 #11

    Dick

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    eumyang marked the problem in red. How did you 1/ln(x-8) in the denominator. That's wrong.
     
  13. Mar 4, 2013 #12
    i'm sorry; it should be.

    (x-9) * (1/(x-8)) + 1(ln(x-8))
     
  14. Mar 4, 2013 #13

    Dick

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    Ok, so you've fixed that. So it's 0/0. Now take another derivative.
     
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