Is the squaring function uniformly continuous on $[0, +\infty)$? Is the function $f: I \to I, ~x \mapsto x^2$ uniformly continuous on $I :=[0, +\infty)$? 
I know that $f$ is uniformly continuous on every interval of the form $[0, b]$, where $b$ is a positive real number. 
 A: No it is not uniformly continuous. Even if $\delta >0$ is very, very small, the difference between $x^2$ and $(x+\delta)^2$ will be large when $x$ is large. When you restrict to $[0,b]$, you're averting this issue by preventing $x$ from being large. 
Here is how you prove it. If $f$ is uniformly continuous, then for any $\epsilon > 0$, there is $\delta > 0$ such that $\lvert x - y \rvert < \delta $ will imply $\lvert f(x) - f(y) \rvert < \epsilon$ for any $x,y$ in the domain (and $\delta$ independent of $x,y$). We'll show that for $\epsilon = 1$, we can never find such $\delta.$
Indeed, for any $\delta > 0$, take $x > 1/(2\delta)$. Then we see $$\lvert f(x) - f(x+\delta) \rvert = 2\delta x + \delta^2 \ge 1 + \delta^2 > 1.$$ Thus we will never have $\lvert f(x) - f(y) \rvert < 1$ for all $x,y \in [0,\infty)$.
Do you see where this would break down if we restrict $x,y \in [0,b]$ for some positive $b$?
A: Take the sequences $x_n=n+\frac{1}{n}$ and $y_n=n$
Then $x_n-y_n =\frac{1}{n}\to 0$  but
$$f(x_n)-f(y_n)=2+\frac{1}{n^2} \to 2 \neq 0$$
Thus $f$ is not uniformly continuous on $[0,+\infty)$
A: Hint: Given any $\epsilon > 0$ and $\delta > 0$, you can always find $x$ large enough such that $|f(x+\delta) - f(x)| > \epsilon$.
A: Yet another way to see: As you note, we are in good shape on intervals of the form $[0,b]$, so let's focus our attention on what happens on intervals of the form $(b,\infty)$. By the mean value theorem,
$$
|f(x)-f(y)|\geq\inf_{c>b}|f'(c)||x-y|=2b|x-y|
$$
for $x,y>b$.
Now, for $\epsilon=1$ and $\delta>0$ whatever you like, we can take $b>\frac{1}{2\delta}$ to push $|f(x)-f(y)|$ above $1$. 
