Finding the Right Delta: Epsilon-Delta Convergence in a Continuous Function

[fyziky]

Hello all,
My question is as follows:
f:[1,\infty) is defined by f(x)=\sqrt{x}+2x (1\leqx<\infty) Given \epsilon>0 find \delta>0 such that if |x-y|<\delta then |f(x)-f(y)|<\epsilon

It seems I am being asked to show continuity, and not uniform continuity, so my approach is this, but I am not sure it works:
|f(x)-f(y)|=\sqrt{x}-\sqrt{y} + 2(x-y) after minor manipulation

since \sqrt{x} function is uniformly continuous it is safe to set \sqrt{x}-\sqrt{y}<\delta/2 and defining |x-y|<\delta/4 we get

\sqrt{x}-\sqrt{y} + 2(x-y) \leq \delta/2 + 2\delta/4=\delta
since \delta can equal \epsilon we just need make |x-y|<\epsilon/4

any help would be appreciated, I am not sure i have the right idea here. thank you.

 

[Raskolnikov]

Yes, if you worded the problem correctly, then this will work because all it's asking is to find some \delta that will work. However, just for kicks and giggles, see if you can find the least upper bound for \delta .

EDIT: On second thought, there's really no need to go that extra step. Nothing really insightful to be gained. Forget it. Your answer's just fine.