How do I prove that $\lim_{(x,y)→(0,0)}\frac{1-\cos(x^2+y^2)}{\sqrt{x^2+y^2}} = 0$ $$\lim_{(x,y)→(0,0)}\frac{1-\cos(x^2+y^2)}{\sqrt{x^2+y^2}} = 0$$
I believe this is correct since I couldn't find a directional limit that won't validate this.
From what I know, I have to prove that
$$\forall\epsilon\gt 0, \exists\delta\gt 0$$
$$ \mbox{such that}$$
$$0 \lt \|(x,y)\| \lt \delta\Rightarrow \frac{1-\cos(x^2+y^2)}{\sqrt{x^2+y^2}} \lt \epsilon$$
I know that $$\sqrt{x^2+y^2} = \|(x,y)\|$$
So I take $$\frac{1-\cos(x^2+y^2)}{\sqrt{x^2+y^2}} \le \frac{2}{\|v\|}$$ and I kind of get stuck there. I greatly appreciate some help.

hint: $$\cos(x^2+y^2) = 1 -\dfrac{(x^2+y^2)^2}{2!}+\dfrac{(x^2+y^2)^4}{4!}-\cdots$$ we are led to the answer as well.
