# Limit of $\frac{x^2-y^2}{x^2+y^2}\sin(x-3y)$ when $(x,y) \to (0,0)$

Show that

$$\lim_{(x,y)\to (0,0)}\frac{x^2-y^2}{x^2+y^2}\sin(x-3y)$$

Does not exists

I've tried the traditional patches, but I always find zero as answer. Any hint?

$$\left|\frac{x^2-y^2}{x^2+y^2}\right|\le 1.$$ So
$$\left|\frac{x^2-y^2}{x^2+y^2}\sin(x-3y)\right|\le |\sin(x-3y)|.$$
$$\lim_{(x,y)\to (0,0)}\sin(x-3y)=0.$$ Thus
$$\lim_{(x,y)\to (0,0)}\left| \frac{x^2-y^2}{x^2+y^2}\sin(x-3y)\right|\le \lim_{(x,y)\to (0,0)}|\sin(x-3y)|=0,$$ from where we conclude that the original limit exists and its value is $0.$