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Evaluate the following limit without using L'Hospital's rule?

  1. Mar 28, 2005 #1
    Can someone please show me, if possible, how to evaluate the following limit without using L'Hospital's rule?

    \mathop {\lim }\limits_{x \to \infty } \left( {1 - \frac{2}{{\log _e n}}} \right)^{\log _e n}

    The answer is e^(-2).

    Rewriting the terms in the bracket as a single term doesn't appear to get me anywhere. I tried taking the logarithm and exponentiating the limit but that still ended up requiring me to use L'Hospital's rule. Any help with evaluating the limit without using L'Hospital's rule would be good thanks.
  2. jcsd
  3. Mar 28, 2005 #2
    This is fairly straightforward...you need to know that

    \mathop {\lim }\limits_{n \to \infty } \left( {1 + \frac{1}{{n}}} \right)^{n} = e

    Hope that helps...

  4. Mar 28, 2005 #3


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    U misstyped it.

    Assuming you're dealing with

    [tex] \lim_{x\rightarrow +\infty}\left(1-\frac{2}{\ln x}\right)^{\ln x} [/tex]

    ,then i advise u to male the obvious substitution

    [tex] \ln x= u [/tex]

    and then use the definition of "e"...

  5. Apr 7, 2005 #4
    It's a bit late, I haven't been online for a while, but thanks for the replies. There might be an error in my booklet but that's exactly(but without the base e, in the subject I'm taking it's assumed that the base is e) how the limit is written in my book.
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