Exploring Limits with Irrational Exponents

[sutupidmath]

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.

 

[Zurtex]

(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 a_n = (-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 a_n^p 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.

 

[sutupidmath]

(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 a_n = (-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 a_n^p 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.

 

[matt grime]

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.

 

[sutupidmath]

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.