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Homework Help: L'Hopitals Rule

  1. Feb 28, 2010 #1
    Evaluate limit

    lim x--> 2+ (5(x-2))^(x-2)

    so what i did i let y= that limit
    Then i took the natural log:

    lim x --> 2+ (x-2) ln| (5(x-2))

    then i put it in the form of L'hopitals rule:

    lim x--> 2+ (ln |(5(x-2))) / 1/(x-2) ...when i plug in 2 for x, both equations DNE so it fits the rule

    So i get the function [5(5x-10)^-1] / [ -(x-2)^-2]
    I just find that by looking at what the equation is, every time i apply l'hopitals rule, it will just keep growing within the negative powers.. im not sure what to do, if i did something wrong, or what i should try?
     
  2. jcsd
  3. Feb 28, 2010 #2

    Dick

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    Try and simplify the algebra before you do l'Hopital again.
     
  4. Feb 28, 2010 #3
    simplify it anymore? like square the denominator?:

    [5/(5x-10)] / [-1/ (x^2-4x+4)]
    thats all i can think off...
     
  5. Feb 28, 2010 #4

    vela

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    You just need to simplify the expression you got after applying L'Hopital's rule, and you'll see nothing blows up when you let x go to 2.
     
  6. Feb 28, 2010 #5

    Dick

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    (x-2) divides (5x-10). There's a more profitable way to simplify.
     
  7. Feb 28, 2010 #6
    oh wait... would i re-raise the equation from the denominator and like re-due l'hopitals rule: here is what i mean if that was confusing

    Here was the equation we had when i first used the rule:
    [5(5x-10)^-1] / [ -(x-2)^-2]

    now ill re raise it
    -(x-2)^2* [5/ (5x-10)]

    and reuse l'hopitals rule but with new equations in the numerator and denominator:
    [ -5(x-2)^2] / [1/ (5x-10)]

    i dont know... maybe im going in the wrong direction.
     
  8. Feb 28, 2010 #7

    Dick

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    Cancel the common factor before you do anything. And your algebra doesn't quite look right there.
     
  9. Feb 28, 2010 #8
    sorry the new equation would be
    [ -5(x-2)^2] / [5x-10]
     
  10. Feb 28, 2010 #9

    Dick

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    You can l'Hopital it again now, or you can just cancel the common factor.
     
  11. Feb 28, 2010 #10
    Oh i think i understand what your saying
    the equation would be:
    5/ (5x-10) * (x-2)^2

    you can simplify the 5x-10 by doing (5(x-2)) So..

    5/5(x-2) * x-2^2

    so the x-2 gets rid off and the sqrd on top gets gone leaving:

    5/5 * x-2 or simply x-2?
     
  12. Feb 28, 2010 #11
    sorry i forgot to include the negative:
    so final answer would be -(x-2) or (2-x)
     
  13. Feb 28, 2010 #12
    so the limit as x--> 2+ for the ln of the function is 0?
     
  14. Feb 28, 2010 #13

    vela

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    Yup.
     
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