Prove that $f: \mathbb{R} \rightarrow \mathbb{R}$, $f(x)=x^k, \forall k \in \{b\in \mathbb{N}|b\geq 2\}$ isn't uniformly continuous.
I believe it isn't uniformly continuous since $x^2$ isn't.
the proof for $k=2$ is:
Let $\epsilon=1$, $\delta>0$, Let $x,y \in \mathbb{R}$ such that $x>y>0, x-y=\delta$ and $x+y>\frac{1}{\delta}$. Then
$$|f(x)-f(y)| = |x^2-y^2| =|x-y||x+y|\geq \delta\cdot\frac{1}{\delta}=1 $$ therefore $x^2$ isn't uniformly continuous.
How do I manage that for any $k>2$?
Assume that $f$ is uniformly continuous. Then we for $\epsilon=1, \exists \delta$, s.t $|x-a|<\delta \implies |f(x) - f(a)|<1$. Obviously this $\delta$ doesn't depend on $x$ nor $a$. Now $\exists n \in \mathbb{N}$, s.t. $n^{k-1}\delta > 1$.
Now choose $a=n, x = n + \frac{\delta}{k}$. Then we have that: $|x-a| = |n + \frac{\delta}{k} - n| = |\frac{\delta}{k}| < \delta \implies 1>|f(n+\frac{\delta}{k}) - f(n)|>|n^k + n^{k-1}\delta - n^k| = |n^{k-1}\delta| > 1$
So we obtain a contradiction.
Let $k>1$, $c>0$ there exists an integer $n$ such that $1/n<c$. Let $x\in R$ such that $kx/n^{k-1}>c$, by using the binomial formula, you obtain you have $|(x+1/n)^k-x^k|\geq kx/n^{k-1}|\geq c$ and $|(x+1/n)-x|=1/n<c$.
