<s>[INST] Find $\lim_{(x,y)\rightarrow(0,0)} \cos(\frac{x^2-y^2}{\sqrt{x^2+y^2}})$ $\lim_{(x,y)\rightarrow(0,0)} \cos(\frac{x^2-y^2}{\sqrt{x^2+y^2}})$
I tried replacing x and y with several values and kept getting 1 so I tried:
$$0 \le |\cos(\frac{x^2-y^2}{\sqrt{x^2+y^2}})| \le |\cos(|(\frac{x^2-y^2}{\sqrt{x^2+y^2}}|)|\le |\cos(\frac{|x|^2+|y|^2}{\sqrt{x^2+y^2}})|\le |\cos(\frac{2\sqrt{x^2+y^2}^2}{\sqrt{x^2+y^2}})| \le |\cos(2\sqrt{x^2+y^2})|$$
Is this correct? I couldn't get Wolfram to compute this properly for some reason.
 [/INST] You can do directly about the $x,y$ :  $ 0 \le \dfrac{x^2}{\sqrt{x^2+y^2}} \le |x|$ and similarly for the other one. This means ...the expression in the $\cos \to 0$ and the answer is clearly $1$.
</s>