Finding a limit of a two variable function: $f(x,y)=\frac {\sin(x^2-xy)}{\vert x\vert} $

Question

I have this exercise but not sure if I'm doing it right $$\lim_{(x,y)\to (0,0)} \frac {\sin(x^2-xy)}{\vert x\vert} $$

I assume $\frac {\sin(x^2-xy)}{\vert x\vert}\le\frac {1}{\vert x \vert} $ then the limit goes to infinite and does not exist.

Is that correct? Or there is some Taylor to use, or some direction to search? Thanks for help.

Answer 1

Your conclusion is not correct, you have just 'proved' that a real number $c$ is such that $c\le \infty$, you can't say that necessarily $c \to \infty$.

One may recall that, as $u \to 0$, $$ |\sin u| \le |u| $$ giving here, as $(x,y) \to (0,0)$, $$ \left|\frac{\sin(x^2-xy)}{|x|}\right|\le \frac{|x^2-xy|}{|x|}=|x-y| \to 0. $$