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Limit evaluation

  1. Nov 27, 2005 #1
    Would anybody solve this problem for me?

    I've tried it for a long time, but don't seem to get the answer.
    I don't think I can apply L'Hospital's rule because the numerator is not zero or indeterminate while the denominator goes to zero

    lim x-> -inf ((1+ e^(1/x))/e^x)

    ok, if I assume the numerator is 1 - e^... and try to solve, I am not able to get rid of the e^x term

  2. jcsd
  3. Nov 27, 2005 #2


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    One of the very first things you should have learned about limits is that if the denominator of a fraction goes to 0 and the numerator does not, then the fraction does not have a limit!
  4. Nov 27, 2005 #3
    I think the answer is [tex] +\infty [/tex] because a number divided by 0 tends to infinity.
  5. Nov 27, 2005 #4


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    Yes, the limit is "infinity" which is just a way of saying that the limit does not exist. It really bothers me to see "a number divided by 0 tends to infinity"! A number cannot be divided by 0 and there is no dividing by 0 in this problem because the denominator is never 0 for any value of x!
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