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\[\lim_{x \rightarrow 0}(\frac{ 1^x + 2^x + 3 ^x ........n^x }{ n })^{\frac{ 1 }{ x }}\]
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Yahoo!
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Power is For Both nume and Deno
hartnn
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can be solved by taking log on both sides...
hartnn
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and also without LH,
do you know what lim x->0 [a^x-1 /x] =... ?
hartnn
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*[a^x-1]/x
Yahoo!
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ln a/b
hartnn
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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 ?
hartnn
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and you need to take log on both sides first....to get 'x' in the denominator.
hartnn
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am i on right track? @experimentX
experimentX
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seems so ... ;let me do on copy!!
experimentX
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oh .. that last one is +1
hartnn
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yep, typo...
hartnn
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\[\lim_{x \rightarrow 0}(\frac{ 1^x + 2^x + 3 ^x ........n^x }{ n })^{\frac{ 1 }{ x }}\\=\lim_{x \rightarrow 0}(\frac{ 1^x + 2^x + 3 ^x ........n^x +n-n}{ n })^{\frac{ 1 }{ x }} \\ \lim_{x \rightarrow 0}(\frac{ 1^x + 2^x + 3 ^x ........n^x-n }{ n }+1)^{\frac{ 1 }{ x }} \\\lim_{x \rightarrow 0}(\frac{ 1^x-1 + 2^x -1+ 3 ^x-1 ........n^x-1 }{ n }+1)^{\frac{ 1 }{ x }}\]
experimentX
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although ... L'hopital seems a bit faster!!
hartnn
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LH after taking log, right ?
experimentX
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yeah!! your method works just fine as well!!
hartnn
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hmm...much faster! thanks~
experimentX
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oh!! i had been using unnecessary approximations. your method is faster!!
experimentX
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should be n! e^(1/n)
experimentX
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woops!! sorry (n!)^(1/n)
hartnn
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using the standard formula for lim x->0 (1+x)^(1/x) and adjusting the exponents .
experimentX
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the other would be use the usual unusual trick!! x = e^ln(x)
hartnn
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haven't practised that much...would like to see steps, if you don't mind...
hartnn
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why do i feel i have asked this limit Q..... O.o, @Yahoo! where are you ?
Yahoo!
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I am Here..:)
Yahoo!
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i also..Need to know the steps Plzz
experimentX
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|dw:1357499465779:dw|
experimentX
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|dw:1357499559730:dw|
experimentX
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|dw:1357499607470:dw|
experimentX
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that's same as @hartnn 's |dw:1357499736048:dw|
experimentX
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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)
hartnn
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got your steps :)
thank you!
experimentX
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well :)
anonymous
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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
Yahoo!
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|dw:1357578098421:dw|
Yahoo!
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Then ??
experimentX
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|dw:1357578176065:dw|
Yahoo!
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Got it..Thxx