Is the limit of the function $\frac{1}{x^r}$ as $x$ approaches infinity necessarily equal to $0$ if $r$ is irrational?

Question

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?

Answer 1

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 $r\to\infty$

Answer 2

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.