Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question

2 Answers 2

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.

share|improve this answer
1  
Well, $x \mapsto \sin (x^2)$ is bounded, but I think it is not UC. –  Siminore Apr 18 at 11:04
    
You are right, I meant to include that the function needs to be monotone. –  user133631 Apr 18 at 11:14

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.