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My textbook states the following theorem:

If $r>0$ is a rational number, then $$\lim_{x \to \infty}\frac{1}{x^r}=0$$ If $r>0$ is a rational number such that $x^r$ is defined for all $x$, then $$\lim_{x \to -\infty}\frac{1}{x^r}=0$$

I'm comfortable with these statements (and can prove them using the epsilon-delta definition for limits at infinity), but I'm wondering if this theorem can be extended to irrational $r$. I'm confident that it can as $x \to \infty$; however, I'm not sure how to conceptualize a negative number raised to an irrational power, let alone determine if $x^r$ is defined for all $x$ if $r$ is irrational. Does anyone know how to approach this? Is this question beyond the scope of calculus?

RyanC
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  • There are a couple ways to define an irrational power. One is to define it using the power series for $e$. Another is to define $a^b$ as the infinimum of $a^q$, where $q$ is any rational greater than $b$. – paulinho Oct 24 '20 at 04:16
  • To the best of my knowledge $x^{(\sqrt{2})}$ is undefined for any negative $x$. This is because $\sqrt{2}$ is actually the limit of a sequence of fractions. But $x^{(1/a)}$ is undefined (in $\mathbb{R}$) when $a$ is even and $x<0$. So if you express $\sqrt{2}$ as the limit of a sequence of fractions, each of whom, in lowest terms has an even # in the denominator, then you are dead in the water. – user2661923 Oct 24 '20 at 04:20
  • For irrational $r > 0$ then $\lim_{x\to \infty}\frac 1{x^r}$ does indeed $\to 0$ as $x^r:(0,\infty)\to (0,\infty)$ is continuous. Presumably though the text hasn't actually defined what $x^r$ where $r$ is irrational yet. But for $x< 0$ and $r$ irrational $x^r$ is not defined and the question would be meaningless. – fleablood Oct 24 '20 at 04:55
  • Good point, @user2661923, although that isn't really a problem for $r=\sqrt2$, or in fact any $r=\sqrt[k]{n}$. The convergents of the continued fraction of $$\sqrt2=\left\{\frac{3}{2},\frac{7}{5},\frac{17}{12},\frac{41}{29},\frac{99}{70},\frac{239}{169},\dots\right\}$$, but we can convert any $\frac{p}{q}$ with even denominator to $\frac{2q}{p}$. Similarly, $$\sqrt[k]{n}\approx\frac{p}{q}\approx\frac{nq^{k-1}}{p^{k-1}}$$. – PM 2Ring Oct 24 '20 at 05:56
  • @PM2Ring Suppose that you construct two sequences of fractions in lowest terms (e.g. Seq-odd and Seq-even), both of which are well defined and both of which converge to $\sqrt{2}$. Seq-odd only uses odd #'s in the denominator, and Seq-even only uses even #'s. Certainly, Seq-odd and Seq-even are both constructible, and certainly Seq-odd allows $(-3)^{\sqrt{2}}$ to be meaningful in $\mathbb{R}$, But there is no rule (that I know of) that insists that Seq-even can't be used. ...see next comment – user2661923 Oct 24 '20 at 06:12
  • @PM2Ring In my opinion, in order for $(-3)^{\sqrt{2}}$ to be well defined, it must be well defined against **any** possible constructed sequence of fractions, just as $(+3)^{\sqrt{2}}$ is. Since $(-3)^{\sqrt{2}}$ is not meaningful against Seq-even, then (in my opinion) you are dead in the water. – user2661923 Oct 24 '20 at 06:12
  • @user2661923 That's fair enough. I'm also not comfortable that Seq-even doesn't work. I just wanted to show that we can produce a Seq-odd for irrationals of that form. – PM 2Ring Oct 24 '20 at 06:16

2 Answers2

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We can think of $x^r$ for irrational $r$ as $e^{r\ln(x)}$ which is something you may be more comfortable with. Then the above limit can be written as ${1\over e^{r\ln(x)}}$. Taking the limit at $x$ goes to infinity, this goes to zero. So it seems to me that the first one can be extended to Irrational numbers. As for the second one, I don’t think an analogous statement applies since we can’t define $x$ to an irrational power for negative $x$. See Can you raise a number to an irrational exponent? for a far more detailed explanation of how to define $x$ to an irrational power.

R. Maresca
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If $r>0$ is irrational, then $\lfloor r \rfloor<r<\lceil r\rceil$

So $$\frac{1}{x^{\lceil r\rceil}}<\frac{1}{x^r}<\frac{1}{x^{\lfloor r \rfloor}}$$

Thus for the squeeze theorem $\frac{1}{x^r}\to 0$ as $r\to\infty$

Raffaele
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