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A simple limit that causes a lot of headache

  1. Mar 31, 2009 #1
    Hi everyone!

    consider the following limit:


    Since we get [0/0] (by injectivity of both exp and log functions), it smelled to me like de l'Hospital's rule until I began calculating the derivatives. Then I realised it's somehow useless...

    Besides the correct value of the limit, I am also very interested in the point, why the rule is not applicable here :)

    I appreciate every idea or hint you give me :)

    thanks a lot in advance,

  2. jcsd
  3. Mar 31, 2009 #2


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    The rule applies, I think, but it doesn't help you as it keeps giving 0/0.
    You could consider the derivative and some careful estimates show that it is always negative (and exponentially grows in the -infinity direction).
    Or you could use a formal argument after rewriting something like
    [tex]\lim_{x \to \infty} \ln(1 - 1/x) e^{-x} = \lim_{y \to 0} \ln(1 - y) / e^y = e^{\ln(\ln(1 - y))} e^{-y}[/tex]
  4. Mar 31, 2009 #3
    ok, i'll stick to the formal argument. Could you please elaborate on what exactly you do there. It's obviously not the substitution y=1/x, is it, since you should've changed the whole argument of the exp funct. and not just replace its sign? Well, how does it work then?
  5. Mar 31, 2009 #4


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    Whoops, my mistake. You are right, I cannot do that.

    It is intuitively clear that e^{-x} will go to 0 faster than ln(1 - 1/x) will. Perhaps substituting y = 1/x and making a Taylor expansion around y = 0?

    Also note that the derivative argument, if written down correctly, is no less formal.
  6. Mar 31, 2009 #5

    Ben Niehoff

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    As [itex]x \to \infty[/itex], 1/x is small, so the logarithm can be approximated by

    [tex]\ln (1 - \frac1x) \approx - \frac1x[/tex]

    Then it should be clear what is happening in the limit.


    Actually, l'Hopital's should have worked just fine. I get

    [tex]-\lim_{x \to \infty} \frac{e^x}{x-x^2}[/tex]
  7. Mar 31, 2009 #6
    Remember that [tex] ln(1 - 1/x) = - \int_{1 - 1/x}^1 \frac{1}{u} du [/tex]

    So your expression is [tex] -e^x \int_{1 - 1/x}^1 \frac{1}{u} du [/tex]. By the MVT of integral calculus [tex] \int_{1 - 1/x}^1 \frac{1}{u} du = \frac{1}{x} \frac{1}{c} [/tex] where c is some value between 1 - 1/x and 1.

    Your expression in your limit is now [tex] \frac{-e^x}{xc} [/tex]. Since c is some value between 1 - 1/x and 1, we see that as x tends to infinity 1/c tends to 1 and it is easy to see that the rest of the expression then goes to negative infinity.
  8. Mar 31, 2009 #7
    Double Post
  9. Apr 1, 2009 #8
    Ben Niehoff, you're right, l'Hospital works fine here after doing a bit of algebra :), btw, I think you've got a minus sign more/less in your answer, since the answer is -infty :)

    JG89, I find your solution using the MVT very elegant, thanks for posting it :)

    CompuChip, Taylor expansion also works very well :)

    So, thanks once again for the help!
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