Uniform continuity of rational function on interval [1,infinity)

Question

To show that $\dfrac{1}{x}$ is uniformly continuous on $[1,\infty)$, could I establish the connection between $|x-y|<\delta$ and $ \bigg|\dfrac{1}{x} - \dfrac{1}{y}\bigg|<\epsilon$ by using the following sequence to connect them

$$ \bigg|\dfrac{1}{x} - \dfrac{1}{y}\bigg|<\epsilon$$ $$\bigg|\dfrac{y-x}{xy}\bigg|<\epsilon$$ $$\dfrac{|y-x|}{xy}<\epsilon$$ $$|x-y|<xy\epsilon$$

then choosing $\delta= xy\epsilon $ and writing the formal proof? Is there anything wrong with this progression? Is there anything I should take not of?

Thanks in advance

Answer 1

There is something really, really wrong with this. You cannot have $\delta$ depending on $x$ and $y$ in the definition of uniform continuity !

One the most important thing to understand about uniform continuity is that $\delta$ should only depend on the $\epsilon > 0$ chosen, not on $x$ nor $y$.

Here, just notice that since $x,y > 1$, you have $xy > 1$ and so $\frac{1}{xy} < 1$, so choosing $\delta = \epsilon$ will work in the definition of uniform continuity.

Answer 2

You could notice that $$x \mapsto1/x$$

has bounded derivative in $[1, \infty)$ so in this interval is uniform continous.