Proving $ f:(0,1)\to R$ is uniformly continuous, where$ f(x)=1/(x+4)$ for all $x \in (0,1)$

Question

I am not sure how to choose delta with regards to epsilon in this question. In my scratch work i get to $\frac{|y-x|}{(x+4)(y+4)}$, and i dont know what to set this <= ? Hope someone can help!

Answer 1

Choose arbitrary $\varepsilon>0$ and let $\delta = \varepsilon /16$. Then we have:

$$|f(x)-f(y)|={|y-x|\over (x+4)(y+4)}< {1\over 16}|x-y| = \varepsilon$$

since $4<x+4<5$ so ${1\over 5}< {1\over x+4}< {1\over 4}$