Determine whether $f(x,y) = \sin(xy)$ is uniformly continuous on $x,y\geq 0$.
Now aside from the definition, I have some trouble even starting with this exercise. I know $|\sin(t)| \leq 1$ forall $t$, and $\sin(t) \leq t$ for positive $t$, but I am just guessing if these could be of use. I am inclined to think that it is uniformly continuous at least.
More generally, how should one approch uniform continuity? I have seen a lot of different approaches, but cannot seem to find a pattern.
Choose $(x_{1,n},y_{1,n}) = (n, 2\pi + \pi/(2n))$ and $(x_{2,n},y_{2,n}) = (n, 2\pi ).$
Then as $n \to \infty$
$$||(x_{1,n},y_{1,n}) - (x_{2,n},y_{2,n})|| = \frac{\pi}{2n} \to 0,$$
but
$$|\sin(x_{1,n}y_{1,n}) - \sin(x_{2,n}y_{2,n})| = |\sin(2\pi n+ \pi/2) - \sin(2 \pi n)| = 1 \not\to 0.$$
Thus, $\sin(xy)$ is not uniformly continuous on $[0,\infty) \times [0,\infty)$.
The selection of this particular sequence of points was motivated by an attempt to find a bound for proving uniform continuity.
In particular,
$$\begin{align}|\sin(x_1y_1) - \sin(x_2y_2)| &= 2 \left|\sin\left(\frac{x_1y_1-x_2y_2}{2}\right) \right|\left|\cos\left(\frac{x_1y_1+x_2y_2}{2}\right) \right| \\ &\leqslant |x_1y_1 - x_2y_2| \\ &\leqslant |x_1||y_1 - y_2| + |y_2||x_1 -x_2|\end{align}.$$
The bound here fails to be small if $|x_1|$ is sufficiently large, despite the proximity of $(x_1,y_1)$ and $(x_2,y_2)$.