Prove that the function$\ f(x)=\sin(x^2)$ is not uniformly continuous on the domain $\mathbb{R}$. If I want to prove that the function$\ f(x)=\sin(x^2)$ is not uniformly continuous on the domain $\mathbb{R}$, I need to show that:
$\exists\varepsilon>0$ $\forall\delta>0$ $\exists{x,y}\in\mathbb{R}\ : |x-y|<\delta$ and $|\sin(x^2) - \sin(y^2)|\geq\varepsilon$.
So let's take $\varepsilon = 1$. Then I want $|\sin(x^2)-\sin(y^2)|\ge1$. That's the case if $\sin(x^2)=0$ and $\sin(y^2)=\pm1$. Thus $x^2=n\pi$ and $y^2=n\pi + \frac{1}{2}\pi$. Now I'm stuck on expressing x and y, which I want to express in $\delta$, to ensure that $|x-y|<\delta$.
Thanks in advance for any help.
 A: You have chosen $x^2=n\pi$ and $y^2 = n\pi+\frac{\pi}{2}$, so you can take $x=\sqrt{n \pi}$ and $y=\sqrt{n \pi + \frac{\pi}{2}}$.
Then,
$|x-y|=\sqrt{n \pi + \frac{\pi}{2}}-\sqrt{n \pi}=\frac{n\pi + \frac{\pi}{2}-n\pi}{\sqrt{n \pi + \frac{\pi}{2}}+\sqrt{n \pi}}=\frac{\frac{\pi}{2}}{\sqrt{n \pi + \frac{\pi}{2}}+\sqrt{n \pi}}<\frac{2}{2\sqrt{n \pi}}<\frac{1}{\sqrt{n}}$
If $n > \frac{1}{\delta^2}$ then $|x-y|<\delta$ but $|f(x)-f(y)|\geq 1$. The values $x,y$ are very close but $f(x)$ and $f(y)$ are far apart.
Intuitively, for $\epsilon = 1$ there is no $\delta$ that allows you to know $f(x)$ within precision $\epsilon$ if you know $x$ within precision $\delta$. The oscillations in $\sin(x^2)$ get faster and faster, on arbitrarily small intervals the function changes its value from 0 to 1.
Note that if you change the function to $\sin x$ the proof will fail, because taking $x=n\pi$ and $y=n\pi+\frac{\pi}{2}$ does not make $|x-y|<\delta$ for small $\delta$ (indeed, $y-x$ is constant). In fact, $\sin x$ is uniformly continuous.
A: Hint: If $x^2=n\pi$ and $\delta>0$, note that $(x+\delta)^2>x^2+2\delta x$.
To make this $>n\pi +\frac12\pi$, it suffices to have $2\delta x > \frac12 \pi$.
This gives you a condition on $n$.
A: You can also say that $f$ is uniformly continuous if and only if for any $x_n$ and $y_n$ such that $(x_n - y_n)\to 0 $ implies $|f(x_n) - f(y_n)| \to 0$ thus you can choose for example $x_n=n$ and $y_n=n-\frac{1}{n}$, which we see $x_n-y_n$. However we have that $f(x_n)-f(y_n) = 2cos(\frac{...}{2})sin(\frac{2-n^2}{2}) \not\to 0$. So we see that the condition for $f$ being uniformly continuous fails.
