Uniformly continuous

  • Thread starter springo
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1. The problem statement, all variables and given/known data
Is the following uniformly continuous on (1,∞)?
[tex]\int_{x}^{x^2}\: e^{-t^2}\mathrm{d}t[/tex]


2. Relevant equations


3. The attempt at a solution
Quite honestly I don't know where to start. I mean, I'm positive I have to use the theorem that says that if for all ε > 0 there exists δ > 0 so that for all x and y, |x - y| < δ implies |f(x) - f(y)| < ε. But that's all I got...

Thanks for your help.
 

Dick

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If you can show a function has bounded derivative over it's domain, then you've shown it's uniformly continuous. If you don't have that theorem yet, just remember the definition of differentiable. It has the epsilons and deltas you need.
 
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126
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[tex]\frac{\mathrm{d}}{\mathrm{d}x}\left(\int_{x}^{x^2}f(t)\,\mathrm{d}t \right)=2x\cdot e^{-x^4}-e^{-x^2}[/tex]

Since x→2x, x→e-x4 and x→e-x2 are bounded on (1,∞), [tex]2x\cdot e^{-x^4}-e^{-x^2}[/tex] is bounded so it's proved.

Is all OK?
Thanks.

PS: I have a similar problem where you have to say whether this is true or false:
|f'(x)|<= K (constant) for -∞< x < ∞
0 < f(x) <= x2 for x >= 1
Then g(x) = f(x)/x is uniformly continuous on (1,∞).
xf'(x)/x2 is bounded and f(x)/x2 too therefore true.
 
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Dick

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If you've proved everything is bounded (like x*e^(-x^4), which does have to be proved, since it's an infinity*0 type of limit - 2x isn't bounded) and you have a theorem that says 'bounded derivative' -> 'uniformly continuous' and don't have to do an epsilon/delta proof, yes.
 
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Oops, yeah I meant 2x is bounded if you don't consider infinity. It's easy to prove that limit, just set u = x4 and have u1/4/eu. What about the other question?
Thanks.
 

Dick

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The other question was just if that proving it had a bounded derivative is enough to satisfy your homework grader. Only you can answer that.
 
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Thanks a lot... I think I'm ready for tomorrow's exam now (lol)
 

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