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?

Thanks in advance!


Note that

$$\left|\frac{x^2-y^2}{x^2+y^2}\right|\le 1.$$ 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 $0.$


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