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Say, $a_n → \alpha$. ($\alpha \in \mathbb{R}$)

How do I prove that the sequence $\{a_n ^r\}$ converges to $\alpha ^r$ when $r\in \mathbb{R}$? (Only if $\alpha ^r$ can be defined)

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    Is this true? What if $\alpha = 0$ and $r = -1$?2012-09-20
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    How do you define $\alpha^r$ when $\alpha\in\mathbb C$ and $r\in\mathbb R$?2012-09-20
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    Edited! My mistake2012-09-20
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    This follows from the fact the $f(x)=x^r$ is continuous, (except at $x=0$ when $r<0$).2012-09-20
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    @Teddy Is there any direct proof for at least when $r\in \mathbb{Q}$?2012-09-20
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    A complete answer requires the basic definition of $x^r$. You need to construct powers as suprema and infima.2012-09-20
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    @Siminore I have constructed it using exponentiation to the power $q\in \mathbb{Q}$. Still, I have trouble with this problem.2012-09-20

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For $r\in\mathbb Q$, you should consider 3 cases:

  1. $r\in\mathbb N$: use $(\alpha_n)^r-\alpha^r = (\alpha_n - \alpha)\cdot\displaystyle\sum_{k=0}^{r-1}{\alpha_n}^k\alpha^{r-1-k}$
  2. $r=\displaystyle\frac 1b,\ b\in\mathbb N$: use the same backwards for $\sqrt[b]{\alpha_n}-\sqrt[b]{\alpha}$
  3. $r=-1$: use $\displaystyle\frac1{\alpha_n}-\frac1\alpha = \frac{\alpha-\alpha_n}{\alpha_n\alpha}$. Here $\alpha\ne 0$ (hence for almost all $n$, $|\alpha_n|$ is above a positive limit) is needed.

For arbitrary $r\in\mathbb R$, in a sense you have to use continuity, anyway exactly continuity defines $\alpha^r$ for all $r\in\mathbb R$..