Ask the proof of $f(x)=x^2$ is not uniformly continuous on $[0, \infty)$ $f(x)=x^2$ is not uniformly continuous on $[0, \infty)$.
Consider the function $f(x)=x^2$. Fix a point $P \in \mathbb{R}, P>0$ and let $\epsilon >0$. In order to guarantee that $|f(x)-f(P)|<\epsilon$, we must have (for x>0) $|x^2-P^2|<\epsilon$, i.e., $|x-P|<\frac{\epsilon}{x+P}$.
From here, I start not to understand: Since x will range over a neighborhood of P, we see that the required $\delta$ in the definition of continuity cannot be larger than $\epsilon/2P$. In fact the choice $|x-P|<\delta=\epsilon/(2P+1)$ will do the job.
Can anyone tell me where does $\epsilon/2P$ from?
And why $|x-P|<\delta=\epsilon/(2P+1)$ will do the job?

 A: If $x - P = \frac{\epsilon}{2P}$, then $x^2 - P^2 = \epsilon + \frac{\epsilon^2}{4P^2} > \epsilon$. So the neighborhood $(x-\delta, x+\delta)$ should not contain $P+\frac{\epsilon}{2P}$.
A: I think the statement is just poorly written. Intuitively, the $\epsilon$/$\delta$ definition of continuity means that everytime you pick an $\epsilon$, you get a $\delta$ such that every point really close of $P$ (at most $\delta$ : $|x - P| < \delta$ ) has its image really close to $f(P)$ (meaning $|f(x)-f(P)|< \epsilon$), the continuity of $f$ forces the images of nearby $x$ to be really close to the image of $P$.
Back to your statement, it just says that for a given $\epsilon$, if your $x$ and $P$ are farther away than $\epsilon/2P$ then the distance between $f(x)$ and $f(P)$ will be greater than $\epsilon$: The points farther than $\epsilon/2P$ are not close enough to one anothers to make the images at most $\epsilon$ close.
You can check the maths, like @angryavian did. Pick $\epsilon > 0$,if we have
$$ |x-P|> \epsilon/2P $$
It means that
$$ |x^2-P^2| = |(x-P)(x+P)| = |x-P|(x+P)$$
Then
$$|x^2-P^2| = |x-P|(x+P)> \epsilon(x+P)/2P$$
it is quite clear that if $x>P$, $|x^2-P^2| > \epsilon$ the images are further away than we want them (namely at most $\epsilon$). It means $\epsilon/2P$ is too big of a $\delta$ for $\epsilon$.
$|x-P|<\delta=\epsilon/(2P+1)$ does the job because for small $\epsilon$ :
$$ |x-P|<\delta=\epsilon/(2P+1) $$
means
$$|x^2-P^2|<\epsilon(x+P)/(2P+1)<\epsilon\frac{2P+\epsilon/(2P+1)}{2P+1}$$
We obtain for $\epsilon$ small enough
$$|x^2-P^2|<\epsilon$$
meaning the "job" is done
Uniform continuity means there is a $\delta$ that does not depend on $P$. If $f$ is uniformly continuous, the required proximity between $x$ and $P$ can depend only on $\epsilon$ no matter where $P$ is. Intuitively, you can see why it is not true for $x^2$ on $\mathbb{R}$
