Uniform Continuity of h(x)=x3+1 on [1, ∞)

[toddlinsley79]

Homework Statement

Is h(x)=x³+1 uniformly continuous on the set [1,infinity)?

The Attempt at a Solution

Let $$\epsilon$$>0. For each x,y in the set [1,infinity) with |x-y|<$$\delta$$, we would have |(x³+1)-(y³+1)|=|x³-y³|

Now how can I show that this is less than epsilon?

 

[Office_Shredder]

It's really important that the question doesn't say the function is uniformly continuous

 

[toddlinsley79]

Ok so |f(x)-f(y)| = (x³+1)-(y³+1)=x³-y³
Now I'm confused about choosing my delta. I thought I was supposed to choose my delta as epsilon divided by (x³-y³) evaluated at x=1 and y=1 because the boundary is [1,infinity).
However, this would lead to a contradiction because x³-y³=0 and I must choose a delta that is smaller than x³-y³=0 and larger than 0.