Limit evaluation

  • #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


thanks
 

Answers and Replies

  • #2
HallsofIvy
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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!
 
  • #3
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I think the answer is [tex] +\infty [/tex] because a number divided by 0 tends to infinity.
 
  • #4
HallsofIvy
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LinkMage said:
I think the answer is [tex] +\infty [/tex] because a number divided by 0 tends to infinity.
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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