<s>[INST] 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.
 [/INST] Here's yet another way
$$\lim\limits_{(x,y)→(0,0)}\frac{1-\cos\left(x^2+y^2\right)}{\sqrt{x^2+y^2}} $$
Using polar coordinates, we have
$$\lim\limits_{r\to 0^+} \frac{1-\cos\left(r^2\cos^2\phi+r^2\sin^2\phi\right)}{\sqrt{r^2\cos^2\phi+r^2\sin^2\phi}} $$
$$=\lim\limits_{r\to 0^+} \frac{1-\cos\left(r^2\left(\cos^2\phi+\sin^2\phi\right)\right)}{\sqrt{r^2}\sqrt{\cos^2\phi+\sin^2\phi}} $$
$$=\lim\limits_{r\to 0^+} \frac{1-\cos\left(r^2\right)}{r} $$
$$=\left(\lim\limits_{r\to 0^+} r\right)\left(\lim\limits_{r\to 0^+} \frac{1-\cos\left(r^2\right)}{r^2} \right)=0$$
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