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How to prove $x^{n}$ is not uniformly continuous on the interval $[0, +\infty)$?

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Presumably you mean for $n>1$. –  Brian M. Scott Aug 12 '12 at 7:40
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Have you tried something? Welcome to MSE :) We would be more enclined and happy to answer if you would've shown what you tried. If this is homework please add the homework-tag ; you tagged calculus so I'll assume you know how tags work. What about $n$? Is it an integer? Is it $\ge 0$, $\ge 1$, we don't know. –  Patrick Da Silva Aug 12 '12 at 7:40
    
Maybe you can look at a related question: Showing a function is not uniformly continuous –  Martin Sleziak Aug 12 '12 at 7:41
    
Look at $(x+\delta)^n-x^n$. –  copper.hat Aug 12 '12 at 7:45
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n is integer and n > 1. I know how to prove when n = 2. proof it by contradiction when n = 2. f(x) = x^{2} x_{1}=\sqrt{n},x_{2}=\sqrt{n+2},for \varepsilon = 1 I get |x_{1}-x_{2}|=\frac{1}{{\sqrt{n}}+{\sqrt{n+2}}} \to 0 but |f(x_{1})-f(x_{2})| = 2 > 1 but how would I chose the sequence when n > 2? –  Mobius Aug 12 '12 at 8:00

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You want to show that there exists some $\epsilon>0$ such that there is no $\delta>0$ for which $$|x-y|<\delta\implies |x^n-y^n|<\epsilon.$$ Try $\epsilon=1$. Thus you want to show that for any $\delta>0$, there is some pair $x,y\in [0,\infty)$ such that $|x-y|<\delta$ yet $|x^n-y^n|\geq 1$. Let's try letting $x=y+\delta/2$. Then we have $|x-y|=\delta/2<\delta$ and $$|x^n-y^n|=(y+\delta/2)^n-y^n\geq (y+\delta/2)y^{n-1}-y^n=\delta/2\cdot y^{n-1}.$$ Can you find some $y$ such that $\delta/2\cdot y^{n-1}$ is at least $1$?

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Thank you, I know how to do next. –  Mobius Aug 12 '12 at 7:59
    
@Mobius, have you interested(sure you do) upvoiting the given answer? You can do it by clicking on the arrow pointing forward. And welcome to M.SE! –  Salech Alhasov Aug 12 '12 at 8:30

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