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anonymous

  • 5 years ago

I have a question involving limits

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  1. anonymous
    • 5 years ago
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    yeah

  2. anonymous
    • 5 years ago
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    I'm sure you do. . . ! Lol.

  3. anonymous
    • 5 years ago
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    \[ \lim_{h \rightarrow 0 } (1+h)^\ln(1+h)-1/h\]

  4. Yuki
    • 5 years ago
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    shoot it

  5. anonymous
    • 5 years ago
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    the ln(1=h) is all together as a power

  6. anonymous
    • 5 years ago
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    the ln(1+h) rather

  7. anonymous
    • 5 years ago
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    it looks like the definition of a derivative... almost

  8. Yuki
    • 5 years ago
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    is it \[\lim_{h \rightarrow 0} ((1+h)^{\ln(1+h)}-1 )/h\] ?

  9. anonymous
    • 5 years ago
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    someone asked this earlier. but use L'hopital's rule and get an indeterminate form of 0/0 by deriving the numerater and denomenator seperately

  10. anonymous
    • 5 years ago
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    yes

  11. anonymous
    • 5 years ago
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    do that and u'll get ur answer. Tell me what you get cuz I don't wanna just give u the answer

  12. anonymous
    • 5 years ago
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    sounds good to me!

  13. anonymous
    • 5 years ago
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    medal pwease

  14. Yuki
    • 5 years ago
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    lol

  15. anonymous
    • 5 years ago
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    i got it to be 0/0 but i don't know where to go exactly from there

  16. anonymous
    • 5 years ago
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    thats ur answer

  17. anonymous
    • 5 years ago
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    ohhh, that easy haha. thanks pointing me in the right direction!

  18. anonymous
    • 5 years ago
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    0/0 is the answer?

  19. anonymous
    • 5 years ago
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    No, 0/0 is an indeterminate answer. If you get a limit in the form of 0/0, you have to use l'Hopital's rule -- if limit of f(x)/g(x) as x ----> infinity = 0/0 or infinity/infinity, then that same limit equals the limit of f'(x)/g'(x). (The limit of the ratio of two functions giving 0/0 or infinity/infinity = the limit of the ratio of their derivatives).

  20. anonymous
    • 5 years ago
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    looks like its back to the drawing board for me then!

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spraguer (Moderator)
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