Uniform continuity of $f(x)=|x^2-1|$ 
Prove that the function $f:\mathbf{R}\rightarrow\mathbf{R}$ given by $f(x)=|x^2-1|$ is not uniformly continuous on $\mathbf{R}$.

I know that I need to prove that there exists an $\varepsilon>0$ such that for all $\delta>0$ we have $|x-y|<\delta$ but $|f(x)-f(y)|\geqslant \varepsilon$. In my book, most of the time they use $\varepsilon=1$ and they use $y=x+\frac{\delta}{2}$ to disprove these things.
$|f(x)-f(y)|=\left\vert|x^2-1|-|y^2-1|\right\vert$ (when I see this I have a strange feeling that tells me I should use the reverse triangle inequality, but this uses $\leqslant$ instead of $\geqslant$..) So maybe I should use that $f(x)=x^2-1$ for $x,y\geqslant 1$? But then I obtain $|f(x)-f(y)|\leqslant |x-y||x+y|$. Let's take $y=x+\frac{\delta}{2}$, but then $|x-y||x+y|=\frac{\delta}{2}|2x+\frac{\delta}{2}|$. If I try to solve for $x$ to make this bigger than $1$, I get $x=\frac{\delta}{4}-\frac{1}{\delta}$, but this is smaller than $1$.. :(
Could someone give a hint for this problem and provide some general hints for disproving uniform continuity?
 A: 
The negation of uniform continuity :  $\exists \epsilon > 0, \forall \delta > 0, \exists x,y > 0 $ such that $|f(x) - f(y)| > \epsilon$ but $|x-y| < \delta$.

Let $\epsilon = 1$, and $\delta > 0$. We want that $|f(x) - f(y)| > 1$, yet $|x - y| < \delta$ for some choice of points $x,y$.
Let $x,y$ be any two real numbers greater than $1$. Then, $x^2 - 1 > 0$, $y^2-1 > 0$, so $|f(x) - f(y)|$ is just $|x^2 - y^2| = |x+y||x-y|$.
Let $y = x + \mu$, where $|\mu| < \delta$. Then, $|f(x) - f(y)| = (\mu)(2x + \mu)$. Set $\mu = \frac \delta 2$ and $x >  \frac{\frac{2}{\delta} - \mu}{2}$, to see that $|f(x) - f(y)| > 1$, yet $|y-x| < \delta$.
That is, if we choose $x > \frac{\frac 2 \delta - \mu}{2}$ positive, and $y = x + \mu$, where $\mu = \frac \delta 2$, then we see that $|f(x) - f(y)| > 1$ although $|x-y| < \delta$.
Hence, points $x,y$ exist, completing the proof.
NOW, if you want general strategies to find if a candidate function is uniformly continuous or not, here is an excellent lemma:

If $f$ is a uniformly continuous function on $\mathbb R$, then there exist $a,b \in \mathbb R^+$ such that $|f(x)| \leq a|x| + b$ for all $x \in \mathbb R$.

That is, a uniformly continuous function has essentially linear growth. That rules out plenty of examples you will get, such as the one in this example, and other possible monsters like $x \log x$, $x^{3/2}$ and so on. Also, it tells you why $\sqrt x, x, \frac{x}{\log x}$ can be expected to be uniformly continuous, since they all obey this lemma.
A: Consider $\epsilon >0.$ Now, for $x,y>1$ we have that
$$|f(y)-f(x)|=|y^2-x^2|=|y-x||x+y|.$$ Consider $\delta >0$ and write $y=x+\dfrac{\delta}{2}.$ We have that
$$|f(x+\delta/2)-f(x)|=\dfrac{\delta}{2}|2x+\delta/2|\ge \delta x.$$ So, if we consider $$x>\dfrac{\epsilon}{\delta}$$ we are done.
A: Take $\varepsilon=1$. Take $x>1$ (so that $f(x)=x^2-1$) and $y=x+\frac\delta2$, where $\delta$ is an abitrary real number greater than $0$. Then$$\bigl|f(y)-f(x)\bigr|=\left(x+\frac\delta2\right)^2-x^2=\delta x+\frac{\delta^2}4.$$You want to have $\delta x+\frac{\delta^2}4\geqslant1=\varepsilon$. Just take $x\geqslant\max\left\{1,\frac 1\delta\right\}$.
