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Math Help - Limit (Where does f(x) go, when k goes towards infinity?) Answer given, but why?

  1. #1
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    Post Limit (Where does f(x) go, when k goes towards infinity?) Answer given, but why?

    Hey math'ers,

    I have  I_k= \left( \ln  \left( 1+{k}^{-1} \right)  \right) ^{k} , where k goes towards infinity...

    I know the answer is supposed to be ln e = 1.

    I would think  {k}^{-1} goes towards  0 , which means the expression inside  \ln would go towards  1 , so  \ln goes towards  0 and the power of  k would just speed up the process.

    What am I doing wrong?
    Any help is appreciated...

    Simon DK
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  2. #2
    TD!
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    For the problem you posted, the answer is indeed 0, but I think you meant:

    I_k= \ln \left(   \left( 1+{k}^{-1} \right) ^{k} \right)

    Then the expression within ln is the standard limit (or definition) of e, so ln(e)=1.
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  3. #3
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    Hello, Simon!

    TD! is absolutely correct.

    You're expected to know the defintion of e:

    . . \lim_{k\to\infty}\left(1 + \frac{1}{k}\right)^k \;=\;e


    I have: . \lim_{k\to\infty}\left[ \ln  \left( 1+{k}^{-1} \right)   ^ k\right]

    I know the answer is supposed to be \ln e = 1

    We have: . \lim_{k\to\infty}\left[ \ln  \left(1+\frac{1}{k}\right)   ^ k\right]<br />
\;= \;\ln\underbrace{\left[\lim_{k\to\infty} \left( 1+\frac{1}{k}\right)^k\right]}_{\text{this is }e} \;= \;\ln(e) \;=\;1
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  4. #4
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    Smile Thanks a whole lot...

    Hey there,

    I'm sorry, I haven't worked on my math project for a little while as it is pre-exam test season now, so I've not been in time to really look at the answer, but it was a great help - thanks, both of you!

    I furthermore notice a few Swedes in the forum questioning about convergence, which I find absolutely perfect. I'm sitting on the side-line looking through answers. ; o ) It must the season of convergence in Scandinavia?

    Simon DK
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