$f(x) = \frac{5x}{2x-1}$ on $[1,\infty)$
Here's what I've worked through so far:
$$|f(x) - f(y)| = \left|\frac{5x}{2x-1} - \frac{5y}{2y-1}\right| = \left|\frac{5y-5x}{(2x-1)(2y-1)}\right| <\epsilon.$$
Since we know $2x-1 \geq 1$ and $2y-1 \geq 1$ for $x,y \in [1,\infty)$, then it suffices to say $y-x < \frac{\epsilon}{5}$?
Am I on the right track? Can someone show me where to go from here?
Yes, you are right. More formally, using your estimates, you can write: $$\left|\frac{5y - 5x}{(2x - 1)(2y - 1)}\right|\leq \left|5y - 5x\right| = 5|y - x|\leq 5\delta.$$ So, take $$\varepsilon = 5\delta\Rightarrow \delta = \frac{\varepsilon}{5}.$$
Since this estimate doesn't depend on the choice of $x$ and $y$, your continuity is uniform.