Uniform continuity of function v continuity

Question

I need to show that a function $f: [1, \infty) \rightarrow \mathbb{R}, f(x) = x^{-1} $ is uniformly continuous.

If I let $ \delta = \epsilon $ and $ | x - a | < \delta $, $ |f(x) - f(a)| = |x^{-1} - a^{-1}| = |\frac{1}{x} - \frac{1}{a}| = \frac{|a-x|}{|ax|} < \frac{\delta}{|ax|} \le \delta = \epsilon$ as $a$ and $x$ are larger than or equal to one (domain). This demonstrates that $$|f(x) - f(a)| < \epsilon$$ I believe this shows that the function is continuous but I don't know what constitutes uniform continuity, or what I need to show to demonstrate that it is uniformly continuous. Any help would be appreciated.

Answer 1

You have already proved uniform continuity. If your $\delta$ depended on $a$ then you would just have continuity at $a$. But since it is independent of $a$ you have proved uniform continuity.

Answer 2

You just have to write it slightly better. Let $\epsilon>0$ be given, choose $\delta=\epsilon$. Now let $x$ and $y$ be in $[1, \infty )$, doing the same you have done, whenever $|x-y|<\delta,$ we get $| f(x) - f(y)| <\epsilon$. As you probably noticed, this is exactly uniform continuity definition.