I'm trying to figure out if the function $f(x)=\cos(x^2)$ is uniformly continuous in $\mathbb{R^2}$. Looking at it's graph It doesn't look like it is, as its graph oscillates more and more violently between $-1$ and $1$ as $x\rightarrow \infty$.
However I've failed to give an $\epsilon-\delta$ proof of this. I've also tried to pick two sequences $\{x_n\},\{y_n\}$ such that $\{x_n-y_n\} \rightarrow 0$ but $\{f(x_n)-f(y_n)\} \not\rightarrow 0$.
On
Try this out. Let $x=\sqrt{2k\pi}$. We show that if $$|\cos(x^2)-\cos((x+\delta)^2)|<\epsilon$$for some $\epsilon>0$ therefore we can make $\delta$ arbitrarily small by taking $x$ sufficiently large. We have $$|\cos(x^2)-\cos((x+\delta)^2)|=|1-\cos(2x\delta+\delta^2)|=2\sin^2(x\delta+\dfrac{\delta^2}{2})<\epsilon$$which means that $$-\sqrt{\dfrac{\epsilon}{2}}<\sin(x\delta+\dfrac{\delta^2}{2})<\sqrt{\dfrac{\epsilon}{2}}$$or $$-2\sin^{-1}\sqrt{\dfrac{\epsilon}{2}}<2x\delta+\delta^2<2\sin^{-1}\sqrt{\dfrac{\epsilon}{2}}$$by adding up $x^2$ to the sides of the inequality we have$$x^2-2\sin^{-1}\sqrt{\dfrac{\epsilon}{2}}<x^2+2x\delta+\delta^2<x^2+2\sin^{-1}\sqrt{\dfrac{\epsilon}{2}}$$which by rearranging means that $$\delta<\dfrac{2\sin^{-1}\sqrt{\dfrac{\epsilon}{2}}}{x+\sqrt{x^2+2\sin^{-1}\sqrt{\dfrac{\epsilon}{2}}}}$$(the other bound is negative and not considered). So we can make $\delta$ arbitrarily small by taking $x$ sufficiently large.
Hint: the derivative is continuous and its absolute value is not bounded from above.