anonymous
  • anonymous
\[\lim_{x \rightarrow \infty} (a ^{1/x} + a ^{-1/x} / 2)^x\] for solving this problem, what are the rational way?
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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goformit100
  • goformit100
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anonymous
  • anonymous
\[\large\lim_{x\to\infty}\left(\frac{a^{1/x}+a^{-1/x}}{2}\right)^x~~?\]
Loser66
  • Loser66
@SithsAndGiggles is it =0?

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anonymous
  • anonymous
I don't know, I haven't worked it out yet. @iaru2, is this your limit?
anonymous
  • anonymous
@loser66, I don't think it's 0. This is going to involve e^(something), which is never 0.
anonymous
  • anonymous
I already solve this problem taking log, But solving way is dirty, and idk be right or don't be right. So I write this thread I wonder what is clear solution of this. for reference, my answer is 1
anonymous
  • anonymous
Please! appear math master ei
anonymous
  • anonymous
Once again, is this your limit? \(\color{blue}{\text{Originally Posted by}}\) @SithsAndGiggles \[\large\lim_{x\to\infty}\left(\frac{a^{1/x}+a^{-1/x}}{2}\right)^x~~?\] \(\color{blue}{\text{End of Quote}}\)
anonymous
  • anonymous
I don't understand your reply, I solved my university exam for my studying I was stuck in this problem. I just wanna various and accurate solution. do you explain what you say?
anonymous
  • anonymous
What the my limit?
anonymous
  • anonymous
I am asking if this is what you are asking. It's hard to tell what your original question says.
anonymous
  • anonymous
Yes, Sorry I don't write correctly, what you write is my limit
anonymous
  • anonymous
Okay. As \(x\to\infty\), you have \(a^{1/x}\to a^0=1\) and \(a^{-1/x}\to a^0=1\), so your limit becomes \[\large\lim_{x\to\infty}\left(\frac{1+1}{2}\right)^x=\lim_{x\to\infty}1^x\]
anonymous
  • anonymous
Ah one more condition is positive number
anonymous
  • anonymous
Right, that's what I had in mind. Do you see why the limit is 1?
anonymous
  • anonymous
Hum. But your explaination is possiblity of hole infinity mutiply of (1.000000000000000000000000000000000001) is infinty intuition say anwser is 1!! But, more explantion need more
anonymous
  • anonymous
I feel this problem is like \[ \frac{ \infty }{ \infty }\]
anonymous
  • anonymous
I have to disagree with what you said earlier. No hole, because as \(x\to\infty\) you have \(a^{\pm1/x}\to1\). Although \(x\) never really becomes "equal" to infinity, let's suppose it does. Then the numerator of the fraction is 1+1 = 2, and divided by 2 you're left with \(1^x\). Here, no matter what \(x\) is equal to, you will always be left with 1.
anonymous
  • anonymous
in here, counter example \[\lim_{x \rightarrow \infty} (a ^{1/x})^{x} \] which is answer 'a' or '1'?
anonymous
  • anonymous
condition : a is positive number
anonymous
  • anonymous
well I imagined it should hold true for any \(a\) so I picked \(e\) to test it:$$\lim_{x\to\infty}\left(\frac12(a^{1/x}+a^{-1/x})\right)^x=\lim_{x\to\infty}\left(\frac12(e^{1/x}+e^{-1/x})\right)^x=\lim_{x\to\infty}(\cosh(1/x))^x$$
anonymous
  • anonymous
\(\color{blue}{\text{Originally Posted by}}\) @iaru2 in here, counter example \[\lim_{x \rightarrow \infty} (a ^{1/x})^{x} \] which is answer 'a' or '1'? \(\color{blue}{\text{End of Quote}}\) This is hardly the same function. The limit would be \(a\), since for all \(x>0\) you have \(\left(a^{1/x}\right)^x=a^{x/x}=a^1=a\).
anonymous
  • anonymous
now consider $$(\cosh(1/x))^x=\exp(x\log(\cosh(1/x)))$$ and we know \(\cosh(1/x)\sim 1+\frac12 (1/x)^2+O(1/x^4)\) as \(|x|\to\infty\) and \(\log(\cosh(1/x))\sim\frac12(1/x)^2+O(1/x^4)\) so \(x\log(\cosh(1/x))\to0\) and thus \((\cosh(1/x))^x\to1\)
anonymous
  • anonymous
@SithsAndGiggles is entirely correct
anonymous
  • anonymous
Hum, oldrin.bataku's solution is interesting... But, i wanna rigor and clear solution, what if other value isn't 1? where is the better solution?
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goformit100
  • goformit100
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anonymous
  • anonymous
iaru2: you discovered that \(2^{-x}(a^{\pm1/x})^x\) behaves differently in the limit than \(2^{-x}(a^{1/x}+a^{-1/x})^x\)
anonymous
  • anonymous
it is true that:$$\lim_{x\to\infty}(a^{1/x})^x=\lim_{x\to\infty} a=a$$
anonymous
  • anonymous
Thank.. But my purpose is counter example that 1^∞ don't always be 1.
anonymous
  • anonymous
hmm?
anonymous
  • anonymous
well take for example $$\lim_{x\to0}(2^x/3^x)^{1/x}=(2^0/3^0)^\infty=1^\infty$$yet$$\lim_{x\to0}(2^x/3^x)^{1/x}=\lim_{x\to0}[(2/3)^x]^{1/x}=2/3$$

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