# Uniformly Continuous

## 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?

## Answers and Replies

Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
It's really important that the question doesn't say the function is uniformly continuous

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.