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amyna

  • one year ago

ok so for this limit it is an infinity over an infinity. But how do i use l hospitals rule for this since its not in fraction form?

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  1. amyna
    • one year ago
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    |dw:1444587028789:dw|

  2. phi
    • one year ago
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    you could find the limit of the ln of that expression then the limit of the original will be e^(limit)

  3. amyna
    • one year ago
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    so i could do 2x ln x ?

  4. phi
    • one year ago
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    yes and write that as \[ 2\frac{\ln x}{x^{-1}} \]

  5. amyna
    • one year ago
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    ok thats all i needed to know! Thank You!

  6. dan815
    • one year ago
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    probably a better idea to do 2x/(1/lnx)

  7. amyna
    • one year ago
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    ya that seems much easier

  8. dan815
    • one year ago
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    the other way simplfies better :)

  9. amyna
    • one year ago
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    i got 0 so do i do the l hospitals rule again?

  10. dan815
    • one year ago
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    ya thats right now apply e^0= 1

  11. amyna
    • one year ago
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    alrighty!!! Thanks!

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