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![[Post New]](/templates/default/images/icon_minipost_new.gif) 24 Jun 2008 04:46:45 IST
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lt x tends to 0 ((1^x +2^x +......+ n^x)/n)^(a/x)
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Cansomeonepleasetellmethelocationofthespacebar ?
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24 Jun 2008 10:33:19 IST
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this reply: 12 points
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24 Jun 2008 11:31:31 IST
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again ....your answer is right but i'm totally confused with the method!! where did the "e" come from??
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Cansomeonepleasetellmethelocationofthespacebar ?
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24 Jun 2008 11:34:36 IST
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![Result\Rightarrow \lim_{x\to0}f(x)^{g(x)}=e^{g(x)(f(x)-1)}\\\lim_{x\to0}(1+f(x)-1)^{g(x)}\\\lim_{x\to0}(1+[f(x)-1])^{\frac{1}{f(x)-1}.g(x)[f(x)-1]}\\\lim_{x\to0}e^{g(x)[f(x)-1]}\\(applied\;result\;\lim_{x\to0}(1+x)^{\frac{1}{x}}=e)](http://alt2.artofproblemsolving.com/Forum/latexrender/pictures/f/1/0/f10d084e1db3f4aa5865b9ff3e0716e99a623c31.gif)
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24 Jun 2008 12:06:43 IST
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Thank you so much!! But can U please say when this result is valid??
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Cansomeonepleasetellmethelocationofthespacebar ?
« ¤ º NoRmaL PeOplE ScAre Me º ¤ »
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24 Jun 2008 12:13:28 IST
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24 Jun 2008 21:25:43 IST
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yes i m also getting the same ans
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25 Jun 2008 14:25:16 IST
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the formula whihc thedumbheadwithnobrains has applied is valid wen the function is of the form 1^infinity....
im talkin abt the formula used to solve the limit in frst step...
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