Limit in two variables of $\frac{\sin(\sqrt[3]x y)}{xy}$ I have to evaluate this limit
$$\lim_{|(x,y)| \to \infty} \frac{\sin(\sqrt[3]x y)}{xy}$$
I think it's useful to consider the module and the inequality $|\sin t| \leq |t|$ for all $t\in\mathbb{R}$ to prove that the limit is $0$, my approach is
$$0\leq \lim_{|(x,y)| \to \infty} \left|\frac{\sin(\sqrt[3]x y)}{xy}\right|\leq \lim_{|(x,y)| \to \infty} \left|\frac{\sqrt[3]x y}{xy}\right|= \lim_{x \to +\infty} \left|\frac{1}{x^{\frac{2}{3}}}\right|=0$$
So the limit is $0$. Is this correct?
If it is correct, I have some more doubts:
(1) In this step: $$\lim_{|(x,y)| \to \infty} \left|\frac{\sqrt[3]x y}{xy}\right|=\lim_{x \to +\infty} \left|\frac{1}{x^{\frac{2}{3}}}\right|$$
The limit becomes a one variable limits just because the function has lost its depencence by $y$?
(2) If my thought on (1) is right, the one variable limit is such that $x\to+\infty$ because $|(x,y)| \to \infty \Leftrightarrow x\to+\infty \ \text{and}\ y\to+\infty$ so, eliminated the depencence of $y$, $x$ has necessarily to tend to $+\infty$?
(3) Why is $|\sin t| \leq |t|$ for all $t\in\mathbb{R}$ valid for $\sin(\sqrt[3]x y)$ even if we are in $\mathbb{R}^2$?
Thanks for your time and excuse by bad english.
 A: (1) It is true that if a function doesn't depend on a variable, then sending that variable to anything that doesn't affect the function. You can prove this similarly to how you would prove that $\lim_{x \to a} 1 = 1$. 
However, the expression you've actually written is wrong, and your conceptual error lies in (2). $\|(x,y)\| \to \infty$ does not imply that $|x| \to \infty$. For example, take the sequence $(1/n, n)$ as $n \to \infty$to $\infty$ .
In fact, the limit in the question doesn't exist. Consider the path $\{ (x(u), y(u)) = (u^{-3}, u): u\in [1, \infty) \}$. Clearly as $u \to \infty,$ $\|(x(u), y(u))\| \to \infty$. But on this path the function is $$ f(u^{-3},u) = \frac{\sin (1)}{u^{-2}} = u^2 \sin(1),$$ which trivially blows up. Thus, along this path $f$ doesn't have a limit, ergo, the limit doesn't exist. [Note that the limit is not even $+\infty$ - go along the path $(4^3 u^{-3}, u)$ ]
To reiterate, ${x\to \infty, y \to \infty}$ is not the same as ${ \|(x,y)\| \to \infty}.$ If instead we were studying $\lim_{x\to \infty, y \to \infty} f(x,y)$, then your approach would be correct, and yield the limit $0$ (althuogh I would use $|\sin(t)| \le 1$ instead of $\le |t|$, since this is stronger for large $t$.)
For (3), this works because $\sin$ always acts on a single variable. You're defining $t(x,y) = x^{1/3}y,$ which is a map from $\mathbb{R}^2$ to $\mathbb{R}$. Then for this real number $t$, $|\sin(t)| \le |t|$. Then you can go back and substitute the value of $t$ in terms of $x,y$.
