anonymous
  • anonymous
\[\lim_{x \rightarrow 0}(\frac{ 1^x + 2^x + 3 ^x ........n^x }{ n })^{\frac{ 1 }{ x }}\]
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
Power is For Both nume and Deno
hartnn
  • hartnn
can be solved by taking log on both sides...
hartnn
  • hartnn
and also without LH, do you know what lim x->0 [a^x-1 /x] =... ?

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hartnn
  • hartnn
*[a^x-1]/x
anonymous
  • anonymous
ln a/b
hartnn
  • hartnn
\[\lim_{x \rightarrow 0}(\frac{ 1^x + 2^x + 3 ^x ........n^x }{ n })^{\frac{ 1 }{ x }}\\=\lim_{x \rightarrow 0}(\frac{ 1^x + 2^x + 3 ^x ........n^x +n-n}{ n })^{\frac{ 1 }{ x }} \\ \lim_{x \rightarrow 0}(\frac{ 1^x + 2^x + 3 ^x ........n^x+n }{ n }-1)^{\frac{ 1 }{ x }} \\\lim_{x \rightarrow 0}(\frac{ 1^x-1 + 2^x -1+ 3 ^x-1 ........n^x-1 }{ n }-1)^{\frac{ 1 }{ x }}\] got the hint ?
hartnn
  • hartnn
and you need to take log on both sides first....to get 'x' in the denominator.
hartnn
  • hartnn
am i on right track? @experimentX
experimentX
  • experimentX
seems so ... ;let me do on copy!!
experimentX
  • experimentX
oh .. that last one is +1
hartnn
  • hartnn
yep, typo...
hartnn
  • hartnn
\[\lim_{x \rightarrow 0}(\frac{ 1^x + 2^x + 3 ^x ........n^x }{ n })^{\frac{ 1 }{ x }}\\=\lim_{x \rightarrow 0}(\frac{ 1^x + 2^x + 3 ^x ........n^x +n-n}{ n })^{\frac{ 1 }{ x }} \\ \lim_{x \rightarrow 0}(\frac{ 1^x + 2^x + 3 ^x ........n^x-n }{ n }+1)^{\frac{ 1 }{ x }} \\\lim_{x \rightarrow 0}(\frac{ 1^x-1 + 2^x -1+ 3 ^x-1 ........n^x-1 }{ n }+1)^{\frac{ 1 }{ x }}\]
experimentX
  • experimentX
although ... L'hopital seems a bit faster!!
hartnn
  • hartnn
LH after taking log, right ?
experimentX
  • experimentX
yeah!! your method works just fine as well!!
hartnn
  • hartnn
hmm...much faster! thanks~
experimentX
  • experimentX
oh!! i had been using unnecessary approximations. your method is faster!!
experimentX
  • experimentX
should be n! e^(1/n)
experimentX
  • experimentX
woops!! sorry (n!)^(1/n)
hartnn
  • hartnn
using the standard formula for lim x->0 (1+x)^(1/x) and adjusting the exponents .
experimentX
  • experimentX
the other would be use the usual unusual trick!! x = e^ln(x)
hartnn
  • hartnn
haven't practised that much...would like to see steps, if you don't mind...
hartnn
  • hartnn
why do i feel i have asked this limit Q..... O.o, @Yahoo! where are you ?
anonymous
  • anonymous
I am Here..:)
anonymous
  • anonymous
i also..Need to know the steps Plzz
experimentX
  • experimentX
|dw:1357499465779:dw|
experimentX
  • experimentX
|dw:1357499559730:dw|
experimentX
  • experimentX
|dw:1357499607470:dw|
experimentX
  • experimentX
that's same as @hartnn 's |dw:1357499736048:dw|
experimentX
  • experimentX
x = e^ln(x) is the usual trick for types of f(x)^g(x) like ... n->0 (n!)^(1/n^n)
hartnn
  • hartnn
got your steps :) thank you!
experimentX
  • experimentX
well :)
anonymous
  • anonymous
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
anonymous
  • anonymous
|dw:1357578098421:dw|
anonymous
  • anonymous
Then ??
experimentX
  • experimentX
|dw:1357578176065:dw|
anonymous
  • anonymous
Got it..Thxx

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