Uniform continuity proof $f(x)=e^{-x^2}$ Is the function $f(x)=e^{-x^2}$ uniformly continuous on $[0,\infty)$?
I'm fairly sure that it is uniformly continuous but I'm having a lot of trouble proving it using the $\epsilon$-$\delta$ proof.  Any help would be much appreciated!
 A: For an $\epsilon-\delta$ proof, differentiate $e^{-x^2}$ to see that is has bounded derivative on $[0,\infty)$. This will tell you how to get your $\delta$ for any given $\epsilon$.
For a more general result, think of $e^{-x^2}$ as a continuous, monotone, and bounded function on $[0,\infty)$, and see if that relates to uniform continuity.
A: Try to prove the following general result: if $f \colon \mathbb{R} \to \mathbb{R}$ is continuous and $\lim_{x \to \pm\infty} f(x)=0$, then $f$ is uniformly continuous. See: this thread.
A: Assume $|x|\lt |y|. (|y| \lt |x|$ is similar) This gives us $-x^2\gt -y^2$. Given $\delta \gt 0, |x-y|\lt \delta,$ then for $x \ge 0$,
$$|e^{-x^2}-e^{-y^2}|=e^{-y^2}|e^{y^2-x^2}-1|\le e^{-y^2}|e^{(x+y)^2}-1|=e^{-y^2}|e^{(x-y)^2+4xy}-1|$$
$\forall{x,y}, \exists{M \in \mathbb Q}$ such that $M \gt 4xy$. 
$\forall{y \in \mathbb R}, -y^2\le 0$.
Then $$e^{-y^2}|e^{(x-y)^2+4xy}-1| \lt |e^{\delta^2+M}-1|:=\epsilon_+$$
For $x \lt 0$,
$$|e^{-x^2}-e^{-y^2}|=e^{-y^2}|e^{y^2-x^2}-1|\lt e^{-y^2}|e^{(y-x)^2}-1|\lt |e^{\delta^2}-1|:=\epsilon_-$$
Trivially, if $|x|=|y|, |e^{-x^2}-e^{-y^2}|=0\lt \epsilon \; \forall{\epsilon \gt 0}$.
Hence $f(x)=e^{-x^2}$ is uniformly continuous by definition. 
