Limit by polar coordinates: $\frac{2x(1-\cos(x-y))}{x^2+y^2}$ I need to prove that:
$$\lim_{(x,y)\to (0,0)}\frac{2x(1-\cos(x-y))}{x^2+y^2} = 0$$
I have tried using polar coordinates, but I'm not comfortable with my approach. What I did was saying:
$$|f(\rho\cos\theta,\rho\sin\theta)| = \frac{2\cos\theta(1-\cos(\rho(\cos\theta-\sin\theta))}{\rho} \leq \frac{2(1-\cos(\sqrt{2}\rho))}{\rho}$$
and this last expression converges to $0$ when $\rho$ goes to $0$ uniformly in $\theta$. Is this correct? How can I do it better? I'm also looking for a way to solve this limit without using polar coordinates ( the easier, the better) . Any ideas?

Solution without the use of polar coordinates, as asked by OP at the end of their post.
Rewrite $$\begin{aligned}1-\cos(x-y)&=1-\cos x \cos y - \sin x \sin y \\ &=1-\cos x + \cos x (1-\cos y)- \sin x \sin y, \end{aligned}$$
then $$\begin{aligned}\frac{2x(1-\cos(x-y))}{x^2+y^2}&= 2x\cdot \frac{1-\cos x}{x^2} \cdot\frac{x^2}{x^2+y^2}\\ &+ 2x\cos x \cdot \frac{1-\cos y}{y^2}\cdot\frac{y^2}{x^2+y^2}\\&- 2y\cdot \frac{\sin x}{x}\cdot\frac{\sin y}{y}\cdot\frac{x^2}{x^2+y^2}\end{aligned}$$
From where is
$$\lim_{(x,y)\to (0,0)}\frac{2x(1-\cos(x-y))}{x^2+y^2} = 0.$$ 
