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Finding the limit of (1 - 5x)^(3/x) as x->0

  1. Oct 8, 2008 #1
    1. The problem statement, all variables and given/known data
    lim of x->0 from above of [tex](1-5x)^\frac{3}{x}[/tex]


    3. The attempt at a solution

    I get to [tex]y=(1-5x)^\frac{3}{x}[/tex] but I'm stuck here.
     
  2. jcsd
  3. Oct 8, 2008 #2

    Dick

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    Re: limits

    What's the limit x->infinity of (1+1/x)^x? I'm guessing you know that. Can you rearrange your limit into something in that form. Try a change of variable like 1/u=-5x.
     
  4. Oct 8, 2008 #3
    Re: limits

    Ok I get to this part:

    [tex]ln y = \frac{3}{x} ln(1-5x)[/tex]
    I know the next step should be
    [tex] \frac{\frac{3}{1-5x}\frac{d}{dx}(1-5x)}{1}[/tex]

    but why?
     
  5. Oct 8, 2008 #4

    Dick

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    Re: limits

    You are trying to apply l'Hopital's theorem to the log, aren't you? I didn't recognize it. Ok. The log of y is (3/x)*ln(1-5x). Write that as 3*ln(1-5x)/x. That has a 0/0 form. So you can use l'Hopital. It has the form f(x)/g(x) where f(x) and g(x) go to zero. What does l'Hopital say?
     
    Last edited: Oct 8, 2008
  6. Oct 8, 2008 #5
    Re: limits

    You're right, I don't =(

    The limit you gave me, is it infinity?
     
  7. Oct 8, 2008 #6

    Dick

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    Re: limits

    No, it's e. But that's ok. If you can use l'Hopital it's easier than that. I edited my post a lot after I realized that. Reread it, ok?
     
  8. Oct 8, 2008 #7
    Re: limits

    Alright, thanks.

    I got [tex]\frac{\frac{3(-5)}{1-5x}}{x}[/tex] so far. Is that right?
     
  9. Oct 8, 2008 #8

    Dick

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    Re: limits

    According to l'Hopital, the candidate for the limit is f'(x)/g'(x), right? You took the derivative of the numerator. Why didn't you take the derivative of the denominator?
     
  10. Oct 8, 2008 #9
    Re: limits

    Oh, ok. Oops.

    so [tex](1-5x) * (-15)[/tex] ?
     
  11. Oct 8, 2008 #10

    Dick

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    Re: limits

    Right. Now let x->0. What do you get? And remember this is ln(y). So what is y?
     
  12. Oct 8, 2008 #11
    Re: limits

    ahhh OK. So does [tex] y = e^{-15}[/tex] ?
     
  13. Oct 8, 2008 #12

    Dick

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    Re: limits

    Yes. Now, if you want, tell me as an exercise why the limit I originally asked about, lim x->infinity (1+1/x)^x=e. It's the same thing really. Would be just for fun. If you get that you have it.
     
  14. Oct 8, 2008 #13
    Re: limits

    hmm

    I get to [tex] ln y = x ln (1+\frac{1}{x})[/tex] and I'm stuck. Is that 0/0? If so, how? Or is that infinity over infinity?
     
  15. Oct 8, 2008 #14

    Dick

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    Re: limits

    It's infinity*zero. Change it to ln(1+1/x)/(1/x). Now it's 0/0. Now you can apply l'Hopital. Find f'(x)/g'(x) and let x->infinity. It's easy. You're almost there.
     
  16. Oct 8, 2008 #15
    Re: limits

    Ok. So when you differentiate you get -1 / -1. So then ln y = 1 and y = e^1 = e ?
     
  17. Oct 8, 2008 #16

    Dick

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    Re: limits

    Yes. You win! It's e. Remember the trick that if you see a limit like 0*infinity you can change it into infinity/infinity or 0/0 by moving a term into the numerator or denominator, whichever makes the derivatives easier. Well done.
     
  18. Oct 8, 2008 #17
    Re: limits

    Awesome! Thanks.
    Don't happen to have any more practice problems handy do you? :)

    I have a test tomorrow on l'Hopital, and integration by simple sub, trig sub, partial fractions, and parts. I've been doing the same problems so many times, I'm worried I've just remembered the problems and not the techniques.
     
  19. Oct 9, 2008 #18

    Dick

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    Re: limits

    Nah. Don't have a big pool of practice problems. I'm sure you have plenty. Good luck. You did pretty well tonight once you got your wits about you. Get some sleep.
     
  20. Oct 9, 2008 #19
    Re: limits

    Ok. Thanks again for your help!
     
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