Calc I: Limits at Infinity: Solving r Irrational

[PFuser1232]

http://tutorial.math.lamar.edu/Classes/CalcI/LimitsAtInfinityI.aspx

According to the author, if $c$ is a real number and $r$ is a positive rational number then:
$$\lim_{x →\infty} \frac{c}{x^r} = 0$$
If $x^r$ is defined for $x < 0$ then:
$$\lim_{x →- \infty} \frac{c}{x^r} = 0$$
I understand why $r$ can't be irrational in case two. $x^r$ would not be defined.
However, I can't see why $r$ can't be irrational in case one.

 

[Svein]

However, I can't see why rr can't be irrational in case one.

Well, it can. Rewrite: $\frac{c}{x^{r}}=\frac{c}{e^{r\ln(x)}}$. Then $\lim_{x\rightarrow \infty} \frac{c}{x^{r}}= \lim_{x\rightarrow \infty} \frac{c}{e^{r\ln(x)}}= \lim_{x\rightarrow \infty} e^{\ln (c)-r\ln(x)}$.

 

[HallsofIvy]

Saying "for r rational" does NOT say it is not also true for r irrational. It may or may not be in that case.

 

[PFuser1232]

Saying "for r rational" does NOT say it is not also true for r irrational. It may or may not be in that case.

When is it not true for irrational $r$?

 

[HallsofIvy]

You still do not understand. I am not saying statement is or is not true for any irrational r. I am simply pointing out that the initial statement "According to the author, if c c is a real number and r r is a positive rational number then:
$\lim_{x\to\infty} \frac{c}{x^r}= 0$ does NOT say anything about what happens if r is irrational!