Need to show the following function is uniformly continuous on R Could you please tell me how I am supposed to show that $f(x) = \dfrac{1}{(1+x^2)}$ is uniformly continuous in $\mathbb{R}$.
I did some pre-calculation and found that $|f(x) - f(u)| < \epsilon$ if $\delta = \dfrac{\epsilon}{|x + u|}$
So, is the selection of $\delta$ correct since I am supposed to find one single $\delta > 0$, so is 
$\delta = \dfrac{\epsilon}{|x + u|} > 0$ or not?
 A: A short answer is "use the mean value theorem and conclude that the function is Lipschitz". But this function has interesting properties that one can find in similar problems with less smooth functions, so let's do it from the definition as well.
First let's do some algebra:
$$|f(x)-f(y)| = \left | \frac{1}{1+x^2} - \frac{1}{1+y^2} \right | = \left | \frac{1+y^2-1-x^2}{(1+x^2)(1+y^2)} \right | = \left | \frac{y^2-x^2}{(1+x^2)(1+y^2)} \right |$$
You want to pause here before starting to actually make estimates, because if you're not careful you may wind up estimating too bluntly. The key here is that different things are happening when $x$ is "small" and when $x$ is "large". 
Let $\varepsilon>0$ and consider $M,\delta > 0$ that we will specify later. When $|x| \leq M$, we can simply say that $(1+x^2)(1+y^2) \geq 1$ so the difference is less than $|y^2-x^2|$. (Here we are neglecting $x^2$ and $y^2$ compared to $1$.) Provided you can assume $|x| \leq M$, this is not hard to control. Specifically, assume $\delta < M/2$, then 
$$|y^2-x^2|=|x+y||x-y| \leq 5M|x-y|/2.$$
So there's our estimate for $x \in [-M,M]$.
When $|x| \geq M$, we now have some additional control because the denominator is now large. Specifically, we can again assume $\delta < M/2$ in order to conclude that the denominator is at least $M^4/4$. Then we have an error of at most $4|y^2-x^2|/M^4=4|x+y||x-y|/M^4$. (Here we are neglecting $1$ compared to $x^2$ and $y^2$.) But $|x+y| \leq 5M/2$ again, so our error is at most $10|x-y|/M^3$. 
So now we need to choose $M$ and $\delta$ such that $\delta < M/2$, $5M\delta/2 < \varepsilon$, and $10\delta/M^3 < \varepsilon$. I will leave it to you to piece everything together. (Hint: in this particular case you can take $M = 1$ regardless of $\varepsilon$.)
With slightly more work but no calculus, we can get a Lipschitz constant. In particular, we need to work out what happens if $|x-y| \geq M/2$. In this case we can take the bluntest possible estimate: $|f(x)-f(y)| < 1$. From this we have that for any $M>0$, $|f(x)-f(y)| \leq \min \{ 1,\max \{ 5M|x-y|/2,10|x-y|/M^3 \} \}$. Then you can play with that to get a constant.
A: Uniform continuity means that $\delta $ does not depend on $x$. In this case, the uniform continuity follows from the following fact: if $f$ is continuous and $\lim_{x\to\pm\infty}f(x)=0$, then $f$ is uniformly continuous on $\mathbb R$. 
To prove it, given $\varepsilon>0$ there is $x_0$ such that $|f(x)|<\varepsilon/2$ if $|x|>x_0-1$. By compactness, $f$ is uniformly continuous on $[-x_0,x_0]$; so there exists $\delta_0$ such that if $x,y\in[-x_0,x_0]$ and $|x-y|<\delta_0$, then $|f(x)-f(y)|<\varepsilon$. 
Now let $\delta=\min\{1,\delta_0\}$. So if $|x-y|<\delta$, then either $x,y\in[-x_0,x_0]$ or both $|x|>x_0-1$, $|y|>x_0-1$. In the first case, $|f(x)-f(y)|<\varepsilon$. In the second one, $$|f(x)-f(y)|\leq|f(x)|+|f(y)|<\frac\varepsilon2+\frac\varepsilon2=\varepsilon.$$
A: It is Lipschitz continuous by the mean value theorem (with constant the supremum of the derivative). Note $|\frac{d}{dx}\frac{1}{1+x^2}|\leqslant 1$ on the whole real line $\mathbf{R}$. Every Lipschitz continuous function is uniformly continuous (if $L$ is a Lipschitz constant, for $\varepsilon>0$ take $\delta=\varepsilon/L$, in your problem you may take $L=1$).
