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

Mathematics
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Power is For Both nume and Deno
can be solved by taking log on both sides...
and also without LH, do you know what lim x->0 [a^x-1 /x] =... ?

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Other answers:

*[a^x-1]/x
ln a/b
\[\lim_{x \rightarrow 0}(\frac{ 1^x + 2^x + 3 ^x ........n^x }{ n })^{\frac{ 1 }{ x }}\\=\lim_{x \rightarrow 0}(\frac{ 1^x + 2^x + 3 ^x ........n^x +n-n}{ n })^{\frac{ 1 }{ x }} \\ \lim_{x \rightarrow 0}(\frac{ 1^x + 2^x + 3 ^x ........n^x+n }{ n }-1)^{\frac{ 1 }{ x }} \\\lim_{x \rightarrow 0}(\frac{ 1^x-1 + 2^x -1+ 3 ^x-1 ........n^x-1 }{ n }-1)^{\frac{ 1 }{ x }}\] got the hint ?
and you need to take log on both sides first....to get 'x' in the denominator.
am i on right track? @experimentX
seems so ... ;let me do on copy!!
oh .. that last one is +1
yep, typo...
\[\lim_{x \rightarrow 0}(\frac{ 1^x + 2^x + 3 ^x ........n^x }{ n })^{\frac{ 1 }{ x }}\\=\lim_{x \rightarrow 0}(\frac{ 1^x + 2^x + 3 ^x ........n^x +n-n}{ n })^{\frac{ 1 }{ x }} \\ \lim_{x \rightarrow 0}(\frac{ 1^x + 2^x + 3 ^x ........n^x-n }{ n }+1)^{\frac{ 1 }{ x }} \\\lim_{x \rightarrow 0}(\frac{ 1^x-1 + 2^x -1+ 3 ^x-1 ........n^x-1 }{ n }+1)^{\frac{ 1 }{ x }}\]
although ... L'hopital seems a bit faster!!
LH after taking log, right ?
yeah!! your method works just fine as well!!
hmm...much faster! thanks~
oh!! i had been using unnecessary approximations. your method is faster!!
should be n! e^(1/n)
woops!! sorry (n!)^(1/n)
using the standard formula for lim x->0 (1+x)^(1/x) and adjusting the exponents .
the other would be use the usual unusual trick!! x = e^ln(x)
haven't practised that much...would like to see steps, if you don't mind...
why do i feel i have asked this limit Q..... O.o, @Yahoo! where are you ?
I am Here..:)
i also..Need to know the steps Plzz
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that's same as @hartnn 's |dw:1357499736048:dw|
x = e^ln(x) is the usual trick for types of f(x)^g(x) like ... n->0 (n!)^(1/n^n)
got your steps :) thank you!
well :)
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
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Then ??
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Got it..Thxx

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