We had to first show that $g(x)=\frac{3}{x-5}$ is uniformly continuous on $(2,4.5)$. This I believe I solved correctly, but please let me know if I did anything wrong:
Proof: Let $\epsilon>0$ and choose $\delta=\frac{\epsilon}{12}$. Then $|x-y|<\delta$ implies $|f(x)-f(y)|=|\frac{3}{x-5}-\frac{3}{y-5}|=|\frac{3(y-5)-3(x-5)}{(x-5)(y-5)}|=|\frac{3(y-x)}{(x-5)(y-5)}|=\frac{3|x-y|}{|x-5||y-5|}$. Since $x,y\in{(2,4.5)}$, we have $2<x<4.5$, thus $-3<x-5<-.5$, thus $|x-5|>\frac{1}{2}$, so $\frac{1}{|x-5|}<2$ (and likewise for $y$). Therefore $|f(x)-f(y)|=\frac{3|x-y|}{|x-5||y-5|}<3*2*2*\frac{\epsilon}{12}=\epsilon$. Hence it is uniformly continuous on this interval.
Now I'm asked to show that $g$ is not uniformly continuous on $(2,5)$, which makes sense to me but I'm not exactly sure how to prove it. Here's what I have so far:
I must show $\exists$ $\epsilon>0$ such that $\forall$ $\delta>0$, $\exists$ $x,y \in{(2,5)}$ with $|x-y|<\delta$ and $|f(x)-f(y)|\geq\epsilon$.
It's not much of a start, but I could really use some insight on how to continue and an opinion on my previous proof. Thanks.
For any $\delta>0$, choose some positive integer $N$ large enough such that $\dfrac{1}{N}<\min\left\{\delta,\dfrac{1}{2}\right\}$, then set $x=5-\dfrac{1}{N}$, $y=5-\dfrac{2}{N}$, it is not hard to show that $x,y\in(2,5)$ and that $|x-y|<\dfrac{1}{N}$. Now $|f(x)-f(y)|=\dfrac{3}{2}N>1$.