I'm trying to prove $f: \mathbb{R}_+\to\mathbb{R}$, where $f(x) = x^a$, is continuous for any $a\in\mathbb{R}$, but I can't come up with anything. I would very much prefer an $\varepsilon$-$\delta$ proof, if possible.
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How do you define $x^a$? I mean, for a natural number $a$ it's obvious, even for a rational number it's obvious. But what's $x^\pi$? – Asaf Karagila Dec 08 '13 at 22:20
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A related problem. You need to consider different cases for $a$. – Mhenni Benghorbal Dec 08 '13 at 22:23
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Well, for any $a\in\mathbb{R}$ there is a sequence $(a_i)$ with $a_i\in\mathbb{Q}$ that converges to $a$. So I would define $x^a = \lim_{i\to\infty} x^{a_i}$. – user114158 Dec 08 '13 at 22:25
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@AsafKaragila Can you elaborate on it being obvious for $a\in\mathbb{Q}$? – user114158 Dec 08 '13 at 22:29
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Once you know that $x^{-n}=\frac1{x^n}$ and $x^{\frac1n}=\sqrt[n]x$, and the expected laws of exponentiation make everything clear. – Asaf Karagila Dec 08 '13 at 22:32
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Ah, thought you meant the proof was obvious. But yes, definition is clear. – user114158 Dec 09 '13 at 22:47
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For $x>0$, you have that $x^a=e^{alnx}$ is continuous, as the composition of two continuous functions $e^x$ and $alnx$. When $x<0$ , the definition is more problematic, since you may have expressions like $x^{1/2k}$ for $x<0$, which is not defined for negative numbers.
user99680
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I assume $x>0$, so that's not an issue. However, I would like to do the proof with an $\varepsilon$-$\delta$ argument. – user114158 Dec 08 '13 at 22:29