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anonymous

  • 5 years ago

lim as x approaches infinity (1+3/x)^(5x)

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  1. anonymous
    • 5 years ago
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    1 becuase as x approach infinity the 3/x goes to zero which just leaves 1^(infinity) and 1 to any power is 1

  2. anonymous
    • 5 years ago
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    Okay well it says that the answer is e^15 .. how is that?!

  3. anonymous
    • 5 years ago
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    \[\lim_{x \rightarrow \infty} (1+{3\over x}^{5x})\]is that what it is?

  4. anonymous
    • 5 years ago
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    ops, typo

  5. anonymous
    • 5 years ago
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    the exponent should be out of the parenthesis

  6. anonymous
    • 5 years ago
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    Yes that is right besides the exponent outside of the parenthesis

  7. anonymous
    • 5 years ago
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    hmmm....definitely looks like 1 to me

  8. anonymous
    • 5 years ago
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    Yeah I wasn't sure how they got e^15. I don't understand at all

  9. anonymous
    • 5 years ago
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    the answer is e^15 but i'm not sure how to get there

  10. myininaya
    • 5 years ago
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    e^(5/3)

  11. myininaya
    • 5 years ago
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    i mean e^(15) sorry

  12. anonymous
    • 5 years ago
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    how did yu get there myininaya ?

  13. myininaya
    • 5 years ago
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    recall limit x->infinity (1+1/x)^x=e

  14. anonymous
    • 5 years ago
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    Is that something I should know off the top of my head?

  15. myininaya
    • 5 years ago
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    i guess lol i just recalled it though like a memory or you know lim x->0 sinx/x=1 or some property like that

  16. anonymous
    • 5 years ago
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    oh boy.. i never liked limits anyways.. lol

  17. anonymous
    • 5 years ago
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    good catch myininaya thats just one of those weird math things

  18. anonymous
    • 5 years ago
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    Okay I remember the sinx/x = 1

  19. anonymous
    • 5 years ago
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    So where did the 15 come from

  20. anonymous
    • 5 years ago
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    from the 3/x instead of 1/x and the power of 5x instead of x

  21. myininaya
    • 5 years ago
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    http://answers.yahoo.com/question/index?qid=20110223065841AAHpCSc

  22. myininaya
    • 5 years ago
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    they show that lim x-> infinity (1+a/x)^x=e^a so if we have lim x-> infinity (1+3/x)^x=e^3 and if we have lim x-> infinity (1+3/x)^(5x)=lim x->infinity [(1+3/x)^x]^5= (the inside goes to e^3) = e^[3(5)]=e^15

  23. anonymous
    • 5 years ago
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    Okay that makes sense now

  24. anonymous
    • 5 years ago
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    Thank you so much!

  25. myininaya
    • 5 years ago
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    np

  26. myininaya
    • 5 years ago
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    catherine you can also see if we plug in really big values for x (1+1/x)^x is getting closer to e

  27. anonymous
    • 5 years ago
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    Which is why it comes to e in the above limits?

  28. myininaya
    • 5 years ago
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    yes you do that for any function you can plug values in for whatever x is getting closer to to see where the function is getting closer to as x gets closer to whatever number its approaching

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