Can Limits Inside Irrational Exponents Be Proven?

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SUMMARY

This discussion centers on the proof of limits involving irrational exponents in sequences and functions. The main question posed is whether the limit of a sequence raised to an irrational power can be expressed as the limit of the sequence raised to that power. The consensus is that while the property holds for natural numbers, additional considerations must be made for irrational exponents. Key insights include the use of continuity properties of the exponential and logarithmic functions to establish the validity of the limit transformation.

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  • Knowledge of irrational numbers and their properties
  • Basic concepts of continuity in mathematical functions
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sutupidmath
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limit proof??

well what i am trying to understand,actually proof is if we can get with the limit inside a power (exponent) if the exponent is irrational.
Say we have any sequence (a_n) or any function f(x), let p be irrational then can we do the following, if yes why, if not why?

1. for the sequence

l
\lim_{\substack{\\n\rightarrow \infty}} (a_n)^{p} =(\lim_{\substack{\\n\rightarrow \infty}} a_n)^{p} ?
and
2.\lim_{\substack{\\x\rightarrow x_o}} (f(x))^{p} =(\lim_{\substack{\\x\rightarrow x_o}} f(x))^{p}

I know how to prove this but only when p is from naturals. HOwever i have never come across any such a problem. Only today suddenly this idea crossed my mind, so i thought i might get some suggesstions here.
So what is the proper answer to this??


thnx in advance

P.S if you could tell me where i could find a proof for this, i would be really grateful.
 
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(Sorry, this post isn't planning to add much, I'm just rambling a little to see what I come up with).

So for natrual numbers it's pretty simple assuming that:

\lim_{n \rightarrow \infty} a_na_n = \left( \lim_{n \rightarrow \infty} a_n \right) \left( \lim_{n \rightarrow \infty} a_n \right)

First of all, I want to point out that this may not be true, take the sequence an = (-1)n

So do you have to make a simmilar assumption with p as an irrational? I think so, I also think you have to clearly define what you're talk about when you mean anp where p is irrational, it's probabily wise to go back to that definition and try and build it up from there, clearly stating any assumptions.
 
Zurtex said:
(Sorry, this post isn't planning to add much, I'm just rambling a little to see what I come up with).

So for natrual numbers it's pretty simple assuming that:

\lim_{n \rightarrow \infty} a_na_n = \left( \lim_{n \rightarrow \infty} a_n \right) \left( \lim_{n \rightarrow \infty} a_n \right)

First of all, I want to point out that this may not be true, take the sequence an = (-1)n

So do you have to make a simmilar assumption with p as an irrational? I think so, I also think you have to clearly define what you're talk about when you mean anp where p is irrational, it's probabily wise to go back to that definition and try and build it up from there, clearly stating any assumptions.

I am assuming that the limit of the sequence (a_n) actually exists, when p is natural, but what for example when p is irrational, this is what i am trying to show. let say a_n=(2n-1)/(n+1), so what can we say now for the limit

\lim_{\substack{\\n\rightarrow \infty}} (a_n)^{p} =(\lim_{\substack{\\n\rightarrow \infty}} a_n)^{p}

when p is natural i can clearly go like this, as u stated

\lim_{\substack{\\n\rightarrow \infty}} (a_n)^{p} =\lim_{n \rightarrow \infty} a_na_na_n ...a_n = \left( \lim_{n \rightarrow \infty} a_n \right) \left( \lim_{n \rightarrow \infty} a_n \right)\left( \lim_{n \rightarrow \infty} a_n \right)\left( \lim_{n \rightarrow \infty} a_n \right)...\left( \lim_{n \rightarrow \infty} a_n \right)=(\lim_{\substack{\\n\rightarrow \infty}} a_n)^{p}

But again i cannot figure out how to do it when we have p irrational?? I can clearly not performe the same thing as i did above assuming that p natural.
 
Last edited:
Follow Zurtex's advice: what is x^p for p irrational? It is exp{p logx}, and log is continuous, multiplication by p is continuous, and exp is continuous.
 
matt grime said:
Follow Zurtex's advice: what is x^p for p irrational? It is exp{p logx}, and log is continuous, multiplication by p is continuous, and exp is continuous.

ohoho

I think i need to be much more vigilent in the future. I think i got it now.

Many thanks to u guys.

P.s. If by any chance,in the future, i might encounter any other problems, or need further clarifications, concerning these kind of problems, i will be back. I hope u won't mind.
 

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