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Gg

  • 5 years ago

find limx^(1/(1-e^-x) when x --> 0

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  1. GG
    • 5 years ago
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    \[\lim_{x \rightarrow 0}x ^{1/(1-e ^{-x})}\]

  2. GG
    • 5 years ago
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    \[\lim_{x \rightarrow 0}x ^{1/\ln (1-e ^{-x})}\]

  3. GG
    • 5 years ago
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    the last one is correct

  4. anonymous
    • 5 years ago
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    Take it piece by piece\[e ^{0}\]is what?

  5. GG
    • 5 years ago
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    it's 1

  6. anonymous
    • 5 years ago
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    \[\ln (1-1)\]=What is \[\ln 0\]It is a trick question.

  7. GG
    • 5 years ago
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    \[-\infty\]

  8. GG
    • 5 years ago
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    ?

  9. anonymous
    • 5 years ago
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    Great. I never got that right each time my teacher asked. You are ahead of the game. So put it together, what is the answer?

  10. GG
    • 5 years ago
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    I got x^0, but x goes to 0, so I got 0^0

  11. anonymous
    • 5 years ago
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    Actually 1 over 0 to the power of infinity, I think. Either way our little game ended in a indeterminate form. This is one where you set y equal to your original thingy. Take ln y and take ln of your thingy. You ever did any of those?

  12. GG
    • 5 years ago
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    I don't understand what do you want to say, I am not so good with English. Can you just write solution?

  13. anonymous
    • 5 years ago
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    \[\ln y =\ln x ^{1/\ln(1-e ^{-x)}}\]

  14. GG
    • 5 years ago
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    and now I have to find limes of this? That's all? :)

  15. anonymous
    • 5 years ago
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    Take lim of each side

  16. GG
    • 5 years ago
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    ok :) thank you very much :)

  17. anonymous
    • 5 years ago
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    I forgot: L'hopital

  18. GG
    • 5 years ago
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    what L'hopital ? :)

  19. anonymous
    • 5 years ago
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    L'hopital = 1) derivative, 2) limit. Does not work here.

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