Prove $f(x)=\frac{3x+2}{2x-1}$ is uniformly continuous on $D_r= (-\infty, 1/2 -r] \cup [1/2 +r, \infty)$ I also need to prove that f(x) is not Uniformly continuous on its domain, $D= (-\infty, \frac{1}{2})\cup(\frac{1}{2},\infty)$. I have tried algebreically manipulating |f(x)-f(y)| and obtain $|f(x)-f(y)|=\frac{7|x-y|}{|2x-1||2y-1|}$, but I am not sure how to obtain a delta that doesn't depend on y for $x,y \in D_r$. I am also not sure why it is not uniformly continuous on domain D. 
 A: For the first part, if $x\in D_r$, then either $x\geq 1/2 + r$ or $x\leq 1/2 - r$, that is, either $x-1/2 \geq r$ or $x-1/2\leq -r$. This is equivalent to saying $|x-1/2|\geq r$. Using what you got, we now have
$$|f(x)-f(y)| = \frac{7|x-y|}{|2x-1| |2y-1|} \leq \frac{7|x-y|}{4r^2}.$$ You can now choose $\delta$ that depends only on $\varepsilon$ and $r$.
For the second part, as you said in the comments: it is enough to show that there are sequences $(x_n)$ and $(y_n)$ such that $x_n-y_n\to 0$ but $f(x_n)-f(y_n)\not\to 0$. If you think about it, there is obviously an issue near $1/2$. So, why not choose sequences that converge to $1/2$ one from above and the other from below?
A: I don't know what tools you are allowed to use. But if you can use differentiability, then you can easily prove the first part.
First, if a function is differentiable, and the derivative is bounded, then it is also Lipschitz (this is rather obvious, follows from Lagrange theorem). Then, you can easily prove that a Lipschitz continuous function is uniformly continuous. Now, look at your function and its derivative in the given set, and conclude.
For the second part, argue by contradiction: assume such uniform $\delta$ exist for a given $\varepsilon$. Can you find two points that are closer than $\delta$ but further apart than $\varepsilon$? The answer is yes, provided you get close enough to 1/2...
