How to show that $f(x)=1/x^2$ is uniformly continuous on $x\ge1$? For my homework I have to show that $f(x)=1/x^2$ is uniformly continuous on the interval $[1,∞)$, and I have absolutely no idea what to do. 
Please don't give me the answer, but if you could push me in the right direction, I would really appreciate it. 

First way
You have in general that if $f:[a,\infty [\to\mathbb R$ is continuous and such that $\lim_{x\to\infty }f(x)=\ell\in\mathbb R$, then $f$ is uniformly continuous. Let $\varepsilon>0$ and $M>a$ s.t. $$|f(x)-\ell|<\frac{\varepsilon}{2}$$ if $x\geq M$. Then $f$ is uniformly continuous on $[a,M]$ (continuous on a compact), and if $x,y>M$ 
$$|f(x)-f(y)|\leq |f(x)-\ell|+|f(y)-\ell|<\varepsilon$$
therefore it's also uniformly continuous on $[M,+\infty [$. Therefore if $\varepsilon>0$, there is a $\delta>0$ such that $$|x-y|<\delta\implies |f(x)-f(y)|<\varepsilon$$
for all $x,y\in [a,M]$ or $x,y\geq M$.
If $x<M<y$ are such that $|x-y|<\delta$, then
$$|f(x)-f(y)|\leq \underbrace{|f(x)-f(M)|}_{\leq \varepsilon}+\underbrace{|f(M)-f(y)|}_{\leq \varepsilon}<2\varepsilon.$$ 
The case $y\leq M\leq x$ is the same, what prove the claim.

Second way
Your function is derivable on $[1,\infty [$. By mean value theorem, for all $x,y\geq 1$ there is a $|c_{x,y}-x|\leq |x-y|$ s.t. $$|f(x)-f(y)|= \underbrace{|f'(c_{x,y})|}_{\leq 2}|x-y|\leq 2|x-y|$$
Then, if $\varepsilon>0$, set $\delta=\frac{\varepsilon}{2}$ and you get$$\forall x,y\geq 1,|x-y|<\delta\implies |f(x)-f(y)|<\varepsilon,$$
what prove the claim.

Third way
(after the deleted answer of clarinetist)
$$\left|\frac{1}{x^2}-\frac{1}{y^2}\right|=\left|\frac{x^2-y^2}{x^2y^2}\right|=\frac{x+y}{x^2y^2}|x-y|.$$
But if $x,y\geq 1$, $$\frac{x+y}{x^2y^2}=\frac{1}{xy^2}+\frac{1}{x^2y}\leq 2$$
and thus $$|\frac{1}{x^2}-\frac{1}{y^2}|\leq 2|x-y|.$$
Let $\varepsilon>0$. If you set $\delta=\frac{\varepsilon}{2}$ you get
$$\forall x,y\geq 1, |x-y|<\delta\implies \left|\frac{1}{x^2}-\frac{1}{y^2}\right|<\varepsilon,$$
what prove the claim.
