# Show that a function is uniformly continuous

How do I show that $f=x/(1+x^2)$ is uniformly continuous on $\Bbb{R}$

Here is what I did: I took the derivative of $f$ and the $\lim_{x\to\infty}f'(x)$ and found that it goes to $0$. So the derivative of $f$ is bounded.

So since the derivative of $f$ is bounded, $f$ is considered to be Lipschitz. SO $${|f(x)-f(y)|\over|x-y|}< M\text{ for }M>0$$

or $|f(x)-f(y)<M|x-y|$ whenever $|x-y|<\delta$. So I choose $M=\epsilon/\delta$

Please give me some feedback on this?

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Your question seems incomplete : what are $\delta$ and $\epsilon$ ? – Siméon Dec 6 '12 at 17:47
Your definition of Lipschitz is wrong. Lipschitz continuity implies $|f(x) - f(y)| \leq M|x - y|$ for "some" $M > 0$ where $M$ is called the Lipschitz "constant". Thereby, in your proof you "cannot choose" $M = \epsilon/\delta$. However, remember that you can choose $\epsilon$ and $\delta$. Also, Lipschitz continuity implies uniform continuity, which is what you are ending up proving. – TenaliRaman Dec 6 '12 at 17:49

You have $f'(x) = -\frac{x^2-1}{(x^2+1)^2}$. A quick calculation shows that $|f'(x)| \leq |f'(0)| = 1$. Hence $|f(x)-f(y)| \leq |x-y|$. Let $\epsilon>0$, then if $|x-y|\leq \epsilon$, you have $|f(x)-f(y)| \leq \epsilon$. Hence $f$ is uniformly continuous on $\mathbb{R}$.
so to show that $|f'(x)\le|f'(0)|$. Do I take the limit of $f'(x)$ and this happens because the limit goes to $0$? – Akaichan Dec 6 '12 at 18:37
Well, you know that $\lim_{|x|\to\infty} |f'(x)| = 0$, and $f'(0) = 1$, so $|f'|$ attains its maximum value at some $x$. To find this, differentiate again and set $f''(x) = 0$. This will give $\{0, \pm \sqrt{3}\}$, and a quick evaluation shows that $x=0$ is where the maximum occurs. – copper.hat Dec 6 '12 at 19:01
ok, I have another question, would it be uniformly continuous if $|f(x)-f(y)|\le\epsilon$, I thought it has to be strictly less than (<) – Akaichan Dec 7 '12 at 15:33
Well, choose $\frac{1}{2} \epsilon$ instead. – copper.hat Dec 7 '12 at 16:06