Prove that $f(x)=frac{1}{x}$ is uniformly continuous on $(frac{1}{2},infty)$

Prove that $f(x)=frac{1}{x}$ is uniformly continuous on $(frac{1}{2},infty)$

Attempt:

We must show that for any $epsilon>0$ we can find a $delta > 0$, such that if $|x-a|<delta$ and $|frac{1}{x}-frac{1}{a}|<epsilon$ for all $a in (frac{1}{2},infty)$.

$quad$Scratchwork for $delta$:.

$begin{align}
quad quad |frac{1}{x}-frac{1}{a}|=|frac{a-x}{xa}|&=fbox{$frac{1}{|xa|}$}|x-a|\&=4|x-a|<epsilon \ &Rightarrow|x-a|<frac{epsilon}{4}
end{align}$

$quad quad $ **In order to bound $frac{1}{|xa|}$ (or $frac{1}{xa}$; since $a,x>0$), I used the fact that $x,a>frac{1}{2}$.

$quad quad $Which would result in $frac{1}{|xa|}=frac{1}{xa}=frac{1}{frac{1}{2}*frac{1}{2}}=4$.

$quad$Choose $delta=frac{epsilon}{4}\$

$quad$Proof:.
Let $epsilon>0$ be given and let $delta=frac{epsilon}{4}$. If $|x-a|<delta$ we have:

$begin{align}
quad quad |frac{1}{x}-frac{1}{a}|&=frac{1}{|xa|}|x-a|\&leq4|x-a|\ &<4delta=4frac{epsilon}{4}=epsilon.
end{align}$

$rule{18cm}{.03cm}$

Is it enough to use the fact that that $x,a>frac{1}{2}$, in order to place bounds on the function in order to reach a $delta$? Any guidance or advice to strengthen the integrity of this proof would be greatly appreciated.