Prove $\frac{1}{x}$ is not uniformly continuous on $[0,\infty)$

Question

Prove $\frac{1}{x}$ is not uniformly continuous on $[0,\infty)$

I was wondering regarding this question as an alternate way to prove by first assuming it's continuous, then using the Mean Value Theroem method:

*show that the derivative $\frac{-1}{x^2} \in [-1,0]\leq [0,\infty)$ ... *

would a contradiction naturally arise from this? I do not know that's why I am asking.

Any input is appreciated!

Answer 1

take $\;x_n=\frac {1}{n} $ and

$y_n=\frac {1}{n+1} .$

we have

$$x_n,y_n\in (0,+\infty) ,$$

$$\lim_{n\to+\infty}(y_n-x_n)=0$$ but $$\lim_{n\to+\infty}(f (y_n)-f (x_n))=1\neq 0$$ thus $f $ is not uniformly continuous at $(0,+\infty) $.