Question

\[\lim_{x \rightarrow \infty} (a ^{1/x} + a ^{-1/x} / 2)^x\]

for solving this problem, what are the rational way?

Answer 1

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Answer 2

\[\large\lim_{x\to\infty}\left(\frac{a^{1/x}+a^{-1/x}}{2}\right)^x~~?\]

Answer 3

@SithsAndGiggles is it =0?

Answer 4

I don't know, I haven't worked it out yet. @iaru2, is this your limit?

Answer 5

@loser66, I don't think it's 0. This is going to involve e^(something), which is never 0.

Answer 6

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

Answer 7

Please! appear math master ei

Answer 8

Once again, is this your limit?

\(\color{blue}{\text{Originally Posted by}}\) @SithsAndGiggles
\[\large\lim_{x\to\infty}\left(\frac{a^{1/x}+a^{-1/x}}{2}\right)^x~~?\]
\(\color{blue}{\text{End of Quote}}\)

Answer 9

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?

Answer 10

What the my limit?

Answer 11

I am asking if this is what you are asking. It's hard to tell what your original question says.

Answer 12

Yes, Sorry I don't write correctly, what you write is my limit

Answer 13

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
\[\large\lim_{x\to\infty}\left(\frac{1+1}{2}\right)^x=\lim_{x\to\infty}1^x\]

Answer 14

Ah one more condition is positive number

Answer 15

Right, that's what I had in mind. Do you see why the limit is 1?

Answer 16

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

Answer 17

I feel this problem is like \[ \frac{ \infty }{ \infty }\]

Answer 18

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.

Answer 19

in here, counter example
\[\lim_{x \rightarrow \infty} (a ^{1/x})^{x} \]
which is answer 'a' or '1'?

Answer 20

condition : a is positive number

Answer 21

well I imagined it should hold true for any \(a\) so I picked \(e\) to test it:$$\lim_{x\to\infty}\left(\frac12(a^{1/x}+a^{-1/x})\right)^x=\lim_{x\to\infty}\left(\frac12(e^{1/x}+e^{-1/x})\right)^x=\lim_{x\to\infty}(\cosh(1/x))^x$$

Answer 22

\(\color{blue}{\text{Originally Posted by}}\) @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 \(\left(a^{1/x}\right)^x=a^{x/x}=a^1=a\).

Answer 23

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 \((\cosh(1/x))^x\to1\)

Answer 24

@SithsAndGiggles is entirely correct

Answer 25

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?

Answer 26

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Answer 27

iaru2: you discovered that \(2^{-x}(a^{\pm1/x})^x\) behaves differently in the limit than \(2^{-x}(a^{1/x}+a^{-1/x})^x\)

Answer 28

it is true that:$$\lim_{x\to\infty}(a^{1/x})^x=\lim_{x\to\infty} a=a$$

Answer 29

Thank.. But my purpose is counter example that 1^∞ don't always be 1.

Answer 30

hmm?

Answer 31

well take for example $$\lim_{x\to0}(2^x/3^x)^{1/x}=(2^0/3^0)^\infty=1^\infty$$yet$$\lim_{x\to0}(2^x/3^x)^{1/x}=\lim_{x\to0}[(2/3)^x]^{1/x}=2/3$$