Uniform Continuity of $1/x^3$ on compact intervals. So, I have the function $f : (0,\infty) \to \Bbb R$. With $f(x) = \frac{1}{x^2}$ and I have to show that $f(x)$ is uniformly continuous on the set $[1,\infty)$. Is the following proof correct?
$[1,\infty) = \cup_{n=1}^{\infty} [n,n+1]$. And each of those sets is compact, and $f(x)$ is continuous so by Heine-Cantor it is UC on every compact set, and thus it is UC on the whole union of them. 
Is this coherent/consistent?
 A: Knowing that it is uniformly continuous on a sequence of compact sets isn't sufficient to know that it is uniformly continuous on their union. Consider the fact that $$(0,\infty)=\bigcup_{n=1}^\infty\left[\frac1n,n+1\right]$$ with your given function to see why not.
I recommend proceeding in the $\epsilon$-$\delta$ fashion, instead. Since your function gets less and less steep as you move to the right, it should suffice to find a $\delta$ that works in $[1,2],$ and show that it actually works everywhere in $[1,\infty).$
A: $[1,\infty)$ is not compact! 
I suggest a workaround without Heine–Cantor as follows: A sufficient condition for uniform continuity is that

$f$ is differentiable, $|f'|$ is bounded.

Then (hover the mouse below to see the spoiler) 

 $$|f(x) - f(y)| = \Bigg|\displaystyle \int^y_x f'(t)dt\Bigg|\leq \int^y_x |f'(t)|dt \leq M|x-y|.$$

A: If $f:[a,\infty)\to\mathbb R$ is a continuous function such that $\lim\limits_{x\to\infty}f(x)=0$, then $f$ is uniformly continuous.  
Proof sketch: Given $\varepsilon>0$, choose $M>a$ such that $|f(x)|<\frac{\varepsilon}{2}$ when $x>M$. Then choose $\delta>0$, $\delta<1$ such that $|f(x)-f(y)|<\varepsilon$ whenever $|x-y|<\delta$ and $x$ and $y$ are in the compact interval $[a,M+1]$.  It follows that $|f(x)-f(y)|<\varepsilon$ whenever $|x-y|<\delta$ and $x$ and $y$ are in $[a,\infty)$.

But for this particular function, you could just play around with $\dfrac{1}{x^2}-\dfrac{1}{y^2}$ to see why it is uniformly continuous, and even Lipschitz.  E.g., $$\left|\dfrac{1}{x^2}-\dfrac{1}{y^2}\right|=\left|\dfrac{1}{x}-\dfrac{1}{y}\right|\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\leq \dfrac{|y-x|}{xy}\left(\dfrac{1}{x}+\dfrac{1}{y}\right).$$  See what happens when you use $x,y\geq 1$?  
Added: I just noticed that I got $\dfrac{1}{x^2}$ from the body of the question, but the title says $\dfrac{1}{x^3}$.  I might not update my answer even if the latter is intended, but a similar approach (slightly more complicated) would apply.
