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1/(e^x-1) limit

  1. Nov 7, 2013 #1
    Hello ,

    i need find the value of [tex] 1/(e^x-1) [/tex], when x is small. The answer is 1/x.

    if i use taylor series [tex] 1/(e^x-1) \approx 1/(1+x-1) =1/x [/tex]

    but

    ¿there is a way to obtain the same answer using [tex] \lim_{x \to 0}1/(e^x-1) [/tex] ???
     
  2. jcsd
  3. Nov 7, 2013 #2

    mfb

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    That limit does not exist. You can evaluate $$\lim_{x \to 0} \frac{x}{e^x-1} = 1$$ and use this to get the approximation 1/x.
     
  4. Nov 7, 2013 #3
    thanks

    why the limit does not exist?,

    why $$ \lim_{x \to 0} \frac{x}{e^x-1} = 1$$???.

    Sorry , but i don't understand
     
  5. Nov 7, 2013 #4

    mfb

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    The numerator is constant, the denominator goes to zero. It does not exist in the same way as the limit of 1/x for x -> 0 does not exist.

    Calculate the limit, and you'll see it.
     
  6. Nov 7, 2013 #5

    Mark44

    Staff: Mentor

    Use the full Maclaurin expansion for ex - 1, and then factor x out of the denominator.
     
  7. Nov 7, 2013 #6
    thanks, very thanks

    and

    ¿there is a way to obtain an answer equal to x using

    $$\lim_{x \to \infty} 1/ \Big (exp (1/x) -1 \Big ) $$ ???
     
  8. Nov 7, 2013 #7

    mfb

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    No. If the limit does not exist, it does not matter how you try to calculate it.
     
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