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

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Yahoo!Best ResponseYou've already chosen the best response.0
Power is For Both nume and Deno
 one year ago

hartnnBest ResponseYou've already chosen the best response.2
can be solved by taking log on both sides...
 one year ago

hartnnBest ResponseYou've already chosen the best response.2
and also without LH, do you know what lim x>0 [a^x1 /x] =... ?
 one year ago

hartnnBest ResponseYou've already chosen the best response.2
\[\lim_{x \rightarrow 0}(\frac{ 1^x + 2^x + 3 ^x ........n^x }{ n })^{\frac{ 1 }{ x }}\\=\lim_{x \rightarrow 0}(\frac{ 1^x + 2^x + 3 ^x ........n^x +nn}{ n })^{\frac{ 1 }{ x }} \\ \lim_{x \rightarrow 0}(\frac{ 1^x + 2^x + 3 ^x ........n^x+n }{ n }1)^{\frac{ 1 }{ x }} \\\lim_{x \rightarrow 0}(\frac{ 1^x1 + 2^x 1+ 3 ^x1 ........n^x1 }{ n }1)^{\frac{ 1 }{ x }}\] got the hint ?
 one year ago

hartnnBest ResponseYou've already chosen the best response.2
and you need to take log on both sides first....to get 'x' in the denominator.
 one year ago

hartnnBest ResponseYou've already chosen the best response.2
am i on right track? @experimentX
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
seems so ... ;let me do on copy!!
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
oh .. that last one is +1
 one year ago

hartnnBest ResponseYou've already chosen the best response.2
\[\lim_{x \rightarrow 0}(\frac{ 1^x + 2^x + 3 ^x ........n^x }{ n })^{\frac{ 1 }{ x }}\\=\lim_{x \rightarrow 0}(\frac{ 1^x + 2^x + 3 ^x ........n^x +nn}{ n })^{\frac{ 1 }{ x }} \\ \lim_{x \rightarrow 0}(\frac{ 1^x + 2^x + 3 ^x ........n^xn }{ n }+1)^{\frac{ 1 }{ x }} \\\lim_{x \rightarrow 0}(\frac{ 1^x1 + 2^x 1+ 3 ^x1 ........n^x1 }{ n }+1)^{\frac{ 1 }{ x }}\]
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
although ... L'hopital seems a bit faster!!
 one year ago

hartnnBest ResponseYou've already chosen the best response.2
LH after taking log, right ?
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
yeah!! your method works just fine as well!!
 one year ago

hartnnBest ResponseYou've already chosen the best response.2
hmm...much faster! thanks~
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
oh!! i had been using unnecessary approximations. your method is faster!!
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
should be n! e^(1/n)
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
woops!! sorry (n!)^(1/n)
 one year ago

hartnnBest ResponseYou've already chosen the best response.2
using the standard formula for lim x>0 (1+x)^(1/x) and adjusting the exponents .
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
the other would be use the usual unusual trick!! x = e^ln(x)
 one year ago

hartnnBest ResponseYou've already chosen the best response.2
haven't practised that much...would like to see steps, if you don't mind...
 one year ago

hartnnBest ResponseYou've already chosen the best response.2
why do i feel i have asked this limit Q..... O.o, @Yahoo! where are you ?
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
i also..Need to know the steps Plzz
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
dw:1357499465779:dw
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
dw:1357499559730:dw
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
dw:1357499607470:dw
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
that's same as @hartnn 's dw:1357499736048:dw
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
x = e^ln(x) is the usual trick for types of f(x)^g(x) like ... n>0 (n!)^(1/n^n)
 one year ago

hartnnBest ResponseYou've already chosen the best response.2
got your steps :) thank you!
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
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
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
dw:1357578176065:dw
 one year ago
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