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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

    Staff: Mentor

    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

    Staff: Mentor

    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.
     
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