# How to prove $x^{n}$ is not uniformly continuous

You want to show that there exists some $\epsilon>0$ such that there is no $\delta>0$ for which $$|x-y|<\delta\implies |x^n-y^n|<\epsilon.$$ Try $\epsilon=1$. Thus you want to show that for any $\delta>0$, there is some pair $x,y\in [0,\infty)$ such that $|x-y|<\delta$ yet $|x^n-y^n|\geq 1$. Let's try letting $x=y+\delta/2$. Then we have $|x-y|=\delta/2<\delta$ and $$|x^n-y^n|=(y+\delta/2)^n-y^n\geq (y+\delta/2)y^{n-1}-y^n=\delta/2\cdot y^{n-1}.$$ Can you find some $y$ such that $\delta/2\cdot y^{n-1}$ is at least $1$?