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

Please give me some feedback on this?

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    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.2012-12-06

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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 $\mathbb{R}$.

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    Well, choose $\frac{1}{2} \epsilon$ instead.2012-12-07