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Homework Help: Limit goes to 0

  1. Feb 18, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the limit of [tex] \lim_{x\rightarrow 0} {{\frac{(5^x-1)}{4x}^\frac{1}{x}} [/tex]


    3. The attempt at a solution
    First I said y = that limit above and took the ln of both sides.

    [tex] ln (y) = \lim_{x\rightarrow 0} {ln{{\frac{(5^x-1)}{4x}^\frac{1}{x}}} [/tex]

    But that gives you -infinity divided by 0 which isn't an indeterminate form. What do I do next?
     
    Last edited: Feb 18, 2010
  2. jcsd
  3. Feb 18, 2010 #2
    The power 1/x goes to the whole fraction, not just to the top. Can't figure out how to get latex to do that.
     
  4. Feb 18, 2010 #3

    Mark44

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    Like this:
    [tex] \lim_{x\rightarrow 0} \left({\frac{5^x-1}{4x}\right)^{\frac{1}{x}} [/tex]

    You can see my LaTeX code by double-clicking it.

    Let y = the expression you're taking the limit of -- without the limit. Now take ln of both sides and work from there. Then take the limit of both sides.
     
  5. Feb 18, 2010 #4
    That's what I did in step 3.

    If you take the limit of [tex]
    \lim_{x\rightarrow 0} \left({\frac{5^x-1}{4x}\right)^{\frac{1}{x}}
    [/tex]

    You get the [tex] \frac{ln(0)}{0} [/tex] Which isn't an indeterminate form. And that's where I'm stuck =(
     
  6. Feb 18, 2010 #5

    Dick

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    How did you get to ln(0)/0? Start by just looking at the (5^x-1)/(4x) part. Does that have a limit?
     
  7. Feb 18, 2010 #6

    Mark44

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    Dick,
    I thought of that, but dismissed it, thinking it wasn't a good idea. I was thinking in terms of limits such as this:
    [tex]\lim_{x \to 0} (1 + x) ^{1/x}[/tex]

    The technique you suggest won't work in the case of this problem, but that's because here we arrive at the indeterminate form [tex][1^{\infty}][/tex], but the OP's problem falls into a completely different category, so I think your suggestion is a good one.
     
  8. Feb 18, 2010 #7

    Dick

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    Hi Mark,

    Right. (5^x-1)/(4x) has indeterminant form 0/0. But you can resolve that with l'Hopital. After that, it's not indeterminant at all. I'm just curious how Zhalfirin88 got to log(0)/0.
     
  9. Feb 18, 2010 #8
    Okay, I took the ln of both sides. So you get:
    [tex]
    ln(\left({\frac{5^x-1}{4x}\right)^{\frac{1}{x}})
    [/tex]

    I rewrote that as

    [tex] \frac{1}{x}ln\left({\frac{5^x-1}{4x}\right) [/tex]

    [tex] \frac{1}{x}\left(ln(5^x-1)-ln(4x)\right) [/tex]

    If you substitute 0:
    [tex] \frac{ln(0)-ln(0)}{0} [/tex]
     
  10. Feb 18, 2010 #9

    Dick

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    You can't 'substitute 0'. This is a limit problem. x/x is also undefined if you substitute x=0 but it has a perfectly good limit. Like I said, I think you should work on the limit of (5^x-1)/(4x) first. It has 0/0 FORM. Which means you can use l'Hopital. But the limit is NOT 0/0.
     
  11. Feb 18, 2010 #10
    What about the exponent?
     
  12. Feb 18, 2010 #11

    Dick

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    Don't worry about the exponent until you figure out what the inside is doing. Trust me.
     
  13. Feb 18, 2010 #12
    Sorry. [tex] \frac{ln(5)}{4} [/tex]
     
  14. Feb 18, 2010 #13

    Dick

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    Right. So as x->0 your limit will look like (ln(5)/4)^(1/x). How does that behave? Be a little careful here.
     
  15. Feb 18, 2010 #14
    I don't know what you're asking.
     
  16. Feb 18, 2010 #15

    Dick

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    I'm asking what's the limit of (ln(5)/4)^(1/x) as x->0. It's not indeterminant.
     
  17. Feb 18, 2010 #16
    edit
     
    Last edited: Feb 18, 2010
  18. Feb 18, 2010 #17

    Dick

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    Well, the log of that quantity is (1/x)*ln(ln(5)/4). You can't use l'Hopital, because that's not indeterminant. Suppose x is positive and close to zero? But you should also consider x is negative and close to zero. If it helps you to think a little more clearly, ln(5)/4~0.4.
     
  19. Feb 18, 2010 #18

    vela

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    That'll give you the wrong answer.

    You need to use the method Mark suggested to take the log and finding its limit.
     
  20. Feb 18, 2010 #19
    Well I have to go to work. And it's due tomorrow morning D= Thanks for your help though.
     
  21. Feb 18, 2010 #20

    Dick

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    Why? I thought we were almost there.
     
  22. Feb 18, 2010 #21
    Why? Isn't it obvious that a number that is less than zero (EDIT: I meant between zero and one) raised to a power that approaches infinity will go to zero?

    Doesn't taking the logarithm just confuse the situation more? Why not keep it simple.

    My simple minded approach was to not even use l'Hospital, but to just look up the expansion of [tex] a^x[/tex] which to first order (with small x) is [tex] 1+x\;{\rm ln} a [/tex]. It all falls out pretty quickly from that. Maybe looking up in tables is frowned upon in math, but engineers are often watching the clock and looking for the quick answer.
     
    Last edited: Feb 18, 2010
  23. Feb 18, 2010 #22

    Dick

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    I think you mean a number less than 1. Sure, that's what I thought. But don't forget x can be negative.
     
  24. Feb 18, 2010 #23
    Yes, quite right. :smile:
     
  25. Feb 18, 2010 #24

    vela

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    Yes, that's obvious, but the error occurred before this point, so it doesn't matter what that limit is.

     
  26. Feb 18, 2010 #25

    vela

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    Try Mark's method. The limit comes out to be 5, not 0.

    Edit: Oh, never mind. I brain-farted and was thinking of the limit as [itex]x\rightarrow\infty[/itex].
     
    Last edited: Feb 18, 2010
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