I want to know if $x^2$ is uniformly continuous on $\Bbb{R^+}$. This kind of question Prove that $f(x)=x^2$ is uniformly continuous on any bounded interval. has been asked before but not on $\Bbb{R^+}$.
I know of a Theorem that states that if $f$ is continuous on $[a,\infty)$ and $\lim\limits_{x\to \infty}f(x)$ exists then $f$ is uniformly continuous. Can anyone help me show if $x^2$ is uniformly continuous on $\Bbb{R^+}$ based on this Theorem?
For uniform continuity, you need to show that given $\def\e{\varepsilon}\e>0$, that there exists a $\delta>0$ so $|x-y|<\delta$ implies $$|x^2-y^2|=|x-y|\cdot|x+y|=\delta|x+y|<\e.$$ But the quantity $|x+y|$ is unbounded, even when you assume $|x-y|<\delta$. Just take $y=x+\delta/2$ to see that $\delta|x+y|$ can get arbitrarily large. In particular, you cannot choose $\delta$ so the above holds for all $x,y$ which are with $\delta$ of each other.
