Explain why the following conjecture for $f(x)=[x]+(x-[x])^{[x]}$ is not correct?

Question

Explain why the following conjectures for $f(x)=[x]+(x-[x])^{[x]}$ are not correct ($x\in(0,\infty)$)

Conjecture a: $\lim_{x\rightarrow\infty}(f(x)-[x])=0.$

Conjecture b: $f$ is uniformly continuous

Thanks.

Answer 1

Edit: This answer assumes that $[x]$ is the integer part of $x$, and therefore $x-[x]\geq 0$, $\forall x\geq 0$.

I think $a$ is false. It seems true, since $0<x-[x]<1$, however, it is not uniformly bounded away from 1. To show it is false, it is enough to show that for every $M>0$ there is at least one $x>M$ such that $(x-[x])^{[x]}>a$, with $a\in (0,1)$ fixed.

Let's say $a=1/2$, and let's pick $x = M+9\sum_{k=1}^n 10^{-k}$. This is means that $x-[x]=0.99...9$, with $k$ 9's after the decimal point. How many do I need? Well, I need

$$ \left(9\sum_{k=1}^n 10^{-k}\right)^M>\frac{1}{2}\\ \Rightarrow 9\sum_{k=1}^n 10^{-k} > \frac{1}{\sqrt[M]{2}}\\ \Rightarrow\sum_{k=1}^n 10^{-k} > \frac{1}{9\sqrt[M]{2}}\\ \Rightarrow\frac{1-\frac{1}{10^{n+1}}}{1-\frac{1}{10}}-1>\frac{1}{9\sqrt[M]{2}}\\ \Rightarrow\frac{10}{9}-\frac{1}{9\cdot 10^n}>1+\frac{1}{9\sqrt[M]{2}}\\ \Rightarrow -\frac{1}{10^n} > -1+\frac{1}{\sqrt[M]{2}}\\ \Rightarrow \frac{1}{10^n} < 1-\frac{1}{\sqrt[M]{2}}=\frac{\sqrt[M]{2}-1}{\sqrt[M]{2}}\\ \Rightarrow 10^n > \frac{\sqrt[M]{2}}{\sqrt[M]{2}-1}\\ \Rightarrow n>\log_{10} \frac{\sqrt[M]{2}}{\sqrt[M]{2}-1}. $$

Clearly, the larger is $M$, the larger is the number of 9's you need to keep to stay above $1/2$, but that's not the point. The set of numbers for which $(x-[x])^{[x]}>1/2$ is unbounded.