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How do I show that $cos(x^2)$ is not a uniformly continuous function?

I try to choose some value of $x=\sqrt{n\pi}$ and $y=\sqrt{n\pi/2}$ so that $|x-y|<\delta$ and $|f(x)-f(y)>\epsilon$, but I got stuck when trying to state the relationship of x and y I picked and $\delta$ in order to make sure that it works for all $\delta$

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As a first step, I highly recommend writing down the definition of uniformly continuous and taking its formal negation to discover what you need to show. –  icurays1 Dec 7 '12 at 4:47
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You'd be better off with $y=\sqrt{n\pi+\frac{\pi}{2}}$, and you could use (because $0<x<y$): $|x-y|=y-x=\frac{y^2-x^2}{x+y}<\frac{y^2-x^2}{2x}$. –  Jonas Meyer Dec 7 '12 at 4:50
    
@JonasMeyer I got it! Thank you so much! –  Akaichan Dec 7 '12 at 5:07
    
What's the domain? Continuous functions with compact domains are uniformly continuous, always. –  anonymous Dec 7 '12 at 19:04

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