Proving $\frac{x}{x-1}$ is uniformly continuous on the interval $(2,4)$.
By definition, $\lvert\frac{x}{x-1}- \frac{y}{y-1}\rvert< \epsilon$. As $4 >x,y > 2$ we can state: $\lvert\frac{x}{x-1} - \frac{y}{y-1}\rvert< \lvert\frac{x}{4-1} -\frac{ y}{4-1}\rvert < 1/3|x-y| < \epsilon/3$. Therefore $\delta = \min\{4,\epsilon/3\}.$
Is this the right approach? How do I prove uniform continuity on the upper bound $4$?
Any hints/help is appreciated.
Pick any $\epsilon > 0$. Take $\delta = \epsilon$. Pick any $x, y \in (2,4)$ such that $|x - y| < \delta$.
$$\bigg| \frac{x}{x -1} - \frac{y}{y-1}\bigg| =\bigg| \frac{x-y}{(x -1)(y -1)} \bigg| = \frac{|x - y|}{|x - 1||y -1|} < \frac{\delta}{|x - 1||y -1|} $$ But we have that $x - 1 > 1$ and $y -1> 1$. So that $$\bigg| \frac{x}{x -1} - \frac{y}{y-1}\bigg| < \frac{\delta}{|x - 1||y -1|} < \delta. $$