Show that the function $f(x)=\frac{1}{x}$is uniformly continuous on the set $A:=[a,\infty )$ where a is positive constant please check my proof 
Let $\epsilon >0 \delta >0$ and a is arbitary positive constant in interval
$$|x-a|<\delta \leftrightarrow |f(x)-f(a)|<\epsilon $$
$$\leftrightarrow |\frac{1}{x}-\frac{1}{a}|<\epsilon $$
$$\leftrightarrow |\frac{a-x}{xa}|<\epsilon $$
$$\leftrightarrow \frac{1}{a}|\frac{a-x}{x}|<\epsilon $$
$$|\frac{a-x}{x}|<a\epsilon $$
choose $\delta =a\epsilon $
then 
$|\frac{1}{x}-\frac{1}{a}|<\delta \leftrightarrow |\frac{1}{x}-\frac{1}{a}<a\frac{\epsilon }{a}=\epsilon $
then it is uniformly conitinouos 

More generally:
(1). For $a,b\in \mathbb R,$ if $f:[a,b]\to \mathbb R$ is continuous then $f$ is uniformly continuous on $[a,b].$
(2). Corollary. If $f:[a,\infty)$ is continuous and $\lim_{x\to \infty}f(x)=0$ then $f$ is uniformly continuous on $[a,\infty).$
Proof of Corollary. Given $e>0,$ take $c\geq a$ such that $\forall x\in [c,\infty)\;(|f(x)|<e/2).$ 
Let $b=c+1.$ 
Take $d\in (0,1)$ such that $\forall x,y\in [a,b]\;(|x-y|<d\implies|f(x)-f(y)|<e/2).$
Now for all $x,y\in (b,\infty)$ we have $x,y\in [c,\infty)$ so$|f(x)-f(y)|\leq |f(x)|+|f(y)|<e/2+e/2=e.$ 
And for $a\leq x\leq b<y,$ if $|x-y|<d$ then $$y>x>y-d>b-d=c+1-d>c,$$ so $\{x,y\}\subset [c,\infty),$ so  $|f(x)-f(y)|\leq |f(x)|+|f(y)|<e/2+e/2=e.$
So altogether we have $\forall x,y \in [a,\infty)\;(|x-y|<d\implies |f(x)-f(y)|<e).$
