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Yahoo!

  • 2 years ago

\[\lim_{x \rightarrow 0}(\frac{ 1^x + 2^x + 3 ^x ........n^x }{ n })^{\frac{ 1 }{ x }}\]

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  1. Yahoo!
    • 2 years ago
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    Power is For Both nume and Deno

  2. hartnn
    • 2 years ago
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    can be solved by taking log on both sides...

  3. hartnn
    • 2 years ago
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    and also without LH, do you know what lim x->0 [a^x-1 /x] =... ?

  4. hartnn
    • 2 years ago
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    *[a^x-1]/x

  5. Yahoo!
    • 2 years ago
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    ln a/b

  6. hartnn
    • 2 years ago
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    \[\lim_{x \rightarrow 0}(\frac{ 1^x + 2^x + 3 ^x ........n^x }{ n })^{\frac{ 1 }{ x }}\\=\lim_{x \rightarrow 0}(\frac{ 1^x + 2^x + 3 ^x ........n^x +n-n}{ n })^{\frac{ 1 }{ x }} \\ \lim_{x \rightarrow 0}(\frac{ 1^x + 2^x + 3 ^x ........n^x+n }{ n }-1)^{\frac{ 1 }{ x }} \\\lim_{x \rightarrow 0}(\frac{ 1^x-1 + 2^x -1+ 3 ^x-1 ........n^x-1 }{ n }-1)^{\frac{ 1 }{ x }}\] got the hint ?

  7. hartnn
    • 2 years ago
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    and you need to take log on both sides first....to get 'x' in the denominator.

  8. hartnn
    • 2 years ago
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    am i on right track? @experimentX

  9. experimentX
    • 2 years ago
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    seems so ... ;let me do on copy!!

  10. experimentX
    • 2 years ago
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    oh .. that last one is +1

  11. hartnn
    • 2 years ago
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    yep, typo...

  12. hartnn
    • 2 years ago
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    \[\lim_{x \rightarrow 0}(\frac{ 1^x + 2^x + 3 ^x ........n^x }{ n })^{\frac{ 1 }{ x }}\\=\lim_{x \rightarrow 0}(\frac{ 1^x + 2^x + 3 ^x ........n^x +n-n}{ n })^{\frac{ 1 }{ x }} \\ \lim_{x \rightarrow 0}(\frac{ 1^x + 2^x + 3 ^x ........n^x-n }{ n }+1)^{\frac{ 1 }{ x }} \\\lim_{x \rightarrow 0}(\frac{ 1^x-1 + 2^x -1+ 3 ^x-1 ........n^x-1 }{ n }+1)^{\frac{ 1 }{ x }}\]

  13. experimentX
    • 2 years ago
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    although ... L'hopital seems a bit faster!!

  14. hartnn
    • 2 years ago
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    LH after taking log, right ?

  15. experimentX
    • 2 years ago
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    yeah!! your method works just fine as well!!

  16. hartnn
    • 2 years ago
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    hmm...much faster! thanks~

  17. experimentX
    • 2 years ago
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    oh!! i had been using unnecessary approximations. your method is faster!!

  18. experimentX
    • 2 years ago
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    should be n! e^(1/n)

  19. experimentX
    • 2 years ago
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    woops!! sorry (n!)^(1/n)

  20. hartnn
    • 2 years ago
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    using the standard formula for lim x->0 (1+x)^(1/x) and adjusting the exponents .

  21. experimentX
    • 2 years ago
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    the other would be use the usual unusual trick!! x = e^ln(x)

  22. hartnn
    • 2 years ago
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    haven't practised that much...would like to see steps, if you don't mind...

  23. hartnn
    • 2 years ago
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    why do i feel i have asked this limit Q..... O.o, @Yahoo! where are you ?

  24. Yahoo!
    • 2 years ago
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    I am Here..:)

  25. Yahoo!
    • 2 years ago
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    i also..Need to know the steps Plzz

  26. experimentX
    • 2 years ago
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    |dw:1357499465779:dw|

  27. experimentX
    • 2 years ago
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    |dw:1357499559730:dw|

  28. experimentX
    • 2 years ago
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    |dw:1357499607470:dw|

  29. experimentX
    • 2 years ago
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    that's same as @hartnn 's |dw:1357499736048:dw|

  30. experimentX
    • 2 years ago
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    x = e^ln(x) is the usual trick for types of f(x)^g(x) like ... n->0 (n!)^(1/n^n)

  31. hartnn
    • 2 years ago
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    got your steps :) thank you!

  32. experimentX
    • 2 years ago
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    well :)

  33. satellite73
    • 2 years ago
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    i can't think of another way at all you are going to have to either take the log or rewrite as an exponential (which is really the same thing) because you have a variable in the exponent

  34. Yahoo!
    • 2 years ago
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    |dw:1357578098421:dw|

  35. Yahoo!
    • 2 years ago
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    Then ??

  36. experimentX
    • 2 years ago
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    |dw:1357578176065:dw|

  37. Yahoo!
    • 2 years ago
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    Got it..Thxx

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