# Epsilon, delta convergence

1. Aug 5, 2010

### fyziky

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

It seems im being asked to show continuity, and not uniform continuity, so my approach is this, but im 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, im not sure i have the right idea here. thank you.

2. Aug 5, 2010

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 $$\delta .$$