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Limits problem

  1. May 30, 2012 #1
    1. The problem statement, all variables and given/known data
    [tex]\stackrel{lim}{x→0}[\frac{(1+x)^{1/x}}{e}]^{1/x}[/tex]
    (its the whole term to the power 1/x, can somebody tell me the correct latex code?)


    2. Relevant equations



    3. The attempt at a solution
    I have actually solved the problem but i am trying to find an alternative to do it.
    My attempt:
    (I am dropping the limit word just to make it easier for me to write)
    let
    [tex]y=\stackrel{lim}{x→0}[\frac{(1+x)^{1/x}}{e}]^{1/x}[/tex]
    [tex]lny=\frac{1}{x}[\frac{ln(1+x)-x}{x}][/tex]
    Using the series of ln(1+x), taking out the common factor x^2 and applying the limit, i get
    [tex]lny=\frac{-1}{2}[/tex]
    [tex]y=e^{\frac{-1}{2}}[/tex]

    Is it possible that i can solve it without using the series expansion of ln(1+x)? Also, i am not allowed to use the L'Hôpital's rule.

    Thanks!
     
  2. jcsd
  3. May 30, 2012 #2

    Mark44

    Staff: Mentor

    Instead of using \stackrel, just use \lim_{x → 0}, like this:
    $$ \lim_{x \to 0}[\frac{(1+x)^{1/x}}{e}]^{1/x}$$
    Let y = the expression inside the limit. Don't include the limit at this step.
    The above might be wrong. You have skipped quite a few steps here, so I'm not following what you did.

    What happened to the inner 1/x exponent? What happened to the e in the denominator?
    This really should go in the Calc & Beyond section - I'm moving it there.
     
  4. May 30, 2012 #3
    Quite sure this is correct, as I get the same expression after simplification. Using logarithm properties, 1/e becomes -1 and the expression boils down to this.

    Probably. What I did is,

    [tex]\lim_{x\to0} (1+f(x))^{g(x)} = e^{\lim_{x\to0} f(x)g(x)}[/tex]

    Which is applicable iff g(x) tends to infinity, which it does.

    From this I get,

    [tex]e^{\lim_{x\to0} \frac{(1+x)^{1/x}-e}{ex}}[/tex]

    (Edit : Probably need to zoom to see this one. Here's the exponent part)

    [tex]{\lim_{x\to0} \frac{(1+x)^{1/x}-e}{ex}}[/tex]
    Finding the limit of this function will give you the answer. You can probably do it by general binomial series, but I'm trying to find a way to get this limit without using L'Hospital Rule or those series.
     
    Last edited: May 30, 2012
  5. May 30, 2012 #4
    OP is right:

    [itex]y=lim_{x\rightarrow 0}(\frac{(1+x)^{1/x}}{e})^{1/x}[/itex]

    [itex]ln(y)=ln(lim_{x\rightarrow 0}(\frac{(1+x)^{1/x}}{e})^{1/x})[/itex]

    [itex]ln(y)=lim_{x\rightarrow 0}ln((\frac{(1+x)^{1/x}}{e})^{1/x})[/itex]

    [itex]ln(y)=lim_{x\rightarrow 0}\frac{1}{x}ln(\frac{(1+x)^{1/x}}{e})[/itex]

    [itex]ln(y)=lim_{x\rightarrow 0}\frac{1}{x}(ln((1+x)^{1/x})-ln(e))[/itex]

    [itex]ln(y)=lim_{x\rightarrow 0}\frac{1}{x}(\frac{1}{x}ln(1+x)-1)[/itex]

    [itex]ln(y)=lim_{x\rightarrow 0}\frac{1}{x}(\frac{ln(1+x)-x}{x})[/itex]
     
  6. May 30, 2012 #5
    Use \left and \right before your parentheses and brackets to have them automatically resize. You can click quote to check out the example below:
    $$
    \lim_{x \to 0} \left( \frac{(1+x)^{1/x}}{e} \right)^{1/x}
    $$
    Your solution is correct (I used L'Hopital's Rule), but I don't see a way to do it without using L'Hopital's rule or series. You could also use the generalized binomial series I suppose, but that's no simpler than your series for log.
     
  7. May 30, 2012 #6
    Thanks for the code Mark! :smile:

    Yes, i did skip a few steps as they involved only simple properties that are mentioned by cjc0117 and Infinitum.

    Why this should go in Calc and Beyond? I haven't used any calculus in this question.:confused: (Now i don't want to get the same infraction again)

    Thanks for trying out the problem. I am asking for an alternative method as i tend to forget the series expansion and binomial which serves me no good in my exams. :smile:

    Exactly what i did. :smile:

    Thanks spamiam, i was in need of this. :smile:
     
  8. May 30, 2012 #7

    Mark44

    Staff: Mentor

    Limits are more associated with calculus than with the precalc courses. Also, this type of problem lends itself to L'Hopitals's Rule, which requires the use of derivatives. In addition, in your approach, you used a series representation of ln(1 + x), which is another calculus concept. For all of these reasons, I moved the thread. I figured it was just an honest mistake on your part -- no infraction.
     
  9. May 30, 2012 #8
    Thanks once again for the clarification, i will take care of that from now. :smile:
     
  10. May 30, 2012 #9
    Try using x=1/n , and taking lim n->Infinity
     
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