$\frac {1}{x}$ behaves nicely in that it's monotone and the derivative is monotone also. So on [$a,\infty$] it can be seen that the $\delta$ which will work everywhere is the $\delta_1$ at the end closest to $a $ when we slide the $\epsilon$ bar. So for the same $\epsilon$ which $f(x)$ and $f(y)$ are in near $a$ the range between $x$ and $y$. To find this $\delta_1$ is simple algebra... $\delta_1= \frac{1}{\frac{1}{a}-\epsilon}-a$.
So: Take $\epsilon>0$ Then $\forall x,y \in [a,\infty)$ s.t $|x-y|< \delta_1= \frac{1}{\frac{1}{a}-\epsilon}-a $
$|f(x)-f(y)|<\epsilon$
Is this a good enough proof? Do I have to show algebraically somehow by taking an arbitrary $x$ and $y$ that $\delta_1$ will work?. I would appreciate any feedback on
a) how to make this a good solution
b)If there is a better solution
c)Good tips/tricks on showing functions in general are uniformly continuous or not.