/sci/ - Science & Math

Okay, I have to prove that if [math]\frac{|f(x)|}{x} < M[/math] for [math]f:[1,\infty) \longrightarrow \mathbb{R}[/math], where f is also continuous and monotonic, then f is uniformly continuous. My attempt was as follows:
Let [math]\epsilon > 0[/math]. Then, assume that f is non-decreasing and that [math]x > y[/math]. Then take [math]\delta = \frac{\epsilon}{M} > |x - y|[/math]. [math] f(x) - f(y) = |f(x) - f(y)| < M|x - y| = Mx - My < \epsilon [/math]. Therefore [math]|x - y| < \delta \implies |f(x) - f(y)| < \epsilon[/math]. Since epsilon doesn't depend on x,y then it is uniformly continuous.
This seems wrong (I didn't use continuity of f) but I'm not sure where I went wrong.