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jennisicle Group Title

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  • 6 months ago
  • 6 months ago

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  1. jennisicle Group Title
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    I think the answer is supposed to be zero but how I get there?

    • 6 months ago
  2. jennisicle Group Title
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    I took the deribative of the top and the bottom but I just ended up at (100n^99)/(e^n) and don't know where to go from here.

    • 6 months ago
  3. myininaya Group Title
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    You should see a pattern in the derivatives.

    • 6 months ago
  4. myininaya Group Title
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    you can do l'hospitals again

    • 6 months ago
  5. myininaya Group Title
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    well nevermind kainui did it for you

    • 6 months ago
  6. myininaya Group Title
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    You could have left it. Maybe it is hard for some people to reach that.

    • 6 months ago
  7. Kainui Group Title
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    The point is, L'H rule says the limit of a ratio is the same as the limit of the ratio of their derivatives as well. So we can just see that eventually n^100 will eventually become a constant while the bottom will always be a function of n.

    • 6 months ago
  8. myininaya Group Title
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    If it is hard for you to see the constant kainui got. You can look at a an example with a lesser power

    • 6 months ago
  9. myininaya Group Title
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    Like what would f^(4)(x) look like if we had f(x)=x^5

    • 6 months ago
  10. jennisicle Group Title
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    I kinda understand what kainui said

    • 6 months ago
  11. myininaya Group Title
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    coolness

    • 6 months ago
  12. Kainui Group Title
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    Sorry, I realized I wasn't being cryptic enough, so I went more cryptical. =P

    • 6 months ago
  13. jennisicle Group Title
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    So does the limit even exist?

    • 6 months ago
  14. Kainui Group Title
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    \[\frac{ d^n }{ dx^n }(x^n)=n!\] just for fun -- Yeah, it even exists and it's even zero. @jennisicle

    • 6 months ago
  15. jennisicle Group Title
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    Ah, that's what I thought, thanks!

    • 6 months ago
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