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A limit question

  1. Nov 24, 2007 #1
    i added a file with a question and a solution from my book

    i added in te file some questions about their solution

    its a high resolution picture
    you can zoom on it.

  2. jcsd
  3. Nov 24, 2007 #2

    Gib Z

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    Homework Helper

    I have absolutely NO idea how the working at the beginning of the picture relates to showing [tex]\lim_{x\to 0} (1+x)^{1/x} = e[/tex].

    However the e limit is quite easy to show if you use the nice property of the natural logarithm, [tex]\log_e \lim_{x\to a} f(x) = \lim_{x\to a} \log_e f(x)[/tex]. In other words, you can interchange the order of limits and logs.

    EDIT: O I did forget to mention you might have to use [tex]\ln(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} x^n = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots \quad{\rm for}\quad \left|x\right| \leq 1\quad {\rm unless}\quad x = -1[/tex], and some people may see that method as somewhat circular, depending on what definitions are used, what is already proved etc etc. But it should be fine.
  4. Nov 24, 2007 #3
    i am puzzled too about it

    how did they get "e"???

    and why the solution of just input (pi/4) instead of X doesnt work?
    that way we get (2.41)^0 thats an possible answer to???
    Last edited: Nov 24, 2007
  5. Nov 24, 2007 #4

    Gib Z

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    What I am puzzled about is what that chunk of working is even about? It doesn't seem to relate to the second part, and even that chunk in itself seems to be quite confusing. Just state the original question first please.
  6. Nov 24, 2007 #5
    the question is:

    lim [tg(pi/8 + x) ] ^ (tg 2x )

  7. Nov 24, 2007 #6
    well u cannot actually do that, cuz tan(pi/2) is not 0 but infinity,so any number raised to the power of infinity is actually undefined. so u have to express it in the form

    e^tg2x ln tg(pi/8 +x)

    and then take the limit as x-->pi/4
  8. Nov 24, 2007 #7
    i cant figure out a way to solve this question??
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