Uniformly Continuous

  • #1
toddlinsley79
6
0

Homework Statement


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


The Attempt at a Solution



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

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

Answers and Replies

  • #2
Office_Shredder
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It's really important that the question doesn't say the function is uniformly continuous
 
  • #3
toddlinsley79
6
0
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.
 

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