<s>[INST] Finding $\lim_{(x,y)\rightarrow (0,0)} \frac{\tan(x^2+y^2)}{\arctan(\frac{1}{x^2+y^2})} $ I'm having trouble understanding how the $\displaystyle\lim_{(x,y)\rightarrow (0,0)}  \frac{\tan(x^2+y^2)}{\arctan(\frac{1}{x^2+y^2})}$. I used the product law to set it up as $\displaystyle\frac{\lim_{(x,y)\rightarrow (0,0)}  \tan(x^2+y^2)}{\lim_{(x,y)\rightarrow (0,0)}\arctan(\frac{1}{x^2+y^2})}$ then found $\lim_{(x,y)\rightarrow (0,0)}  \tan(x^2+y^2)=0$ but got caught up on $\lim_{(x,y)\rightarrow (0,0)}\arctan(\frac{1}{x^2+y^2})$. I looked up the solution and $\lim_{(x,y)\rightarrow (0,0)}\arctan(\frac{1}{x^2+y^2})= \pi/2$. Can someone explain how to solve the second limit?
 [/INST] $$\lim\limits_{(x,y)\to (0,0)}  \frac{\tan\left(x^2+y^2\right)}{\arctan\left(\frac{1}{x^2+y^2}\right)}$$
Using polar coordinates, we have
$$\lim\limits_{r\to 0^+} \frac{\tan\left(r^2\cos^2\phi+r^2\sin^2\phi\right)}{\arctan\left(\frac{1}{r^2\cos^2\phi+r^2\sin^2\phi}\right)}$$
$$=\lim\limits_{r\to 0^+} \frac{\tan\left(r^2\left(\cos^2\phi+\sin^2\phi\right)\right)}{\arctan\left(\frac{1}{r^2\left(\cos^2\phi+\sin^2\phi\right)}\right)}$$
$$=\lim\limits_{r\to 0^+} \frac{\tan\left(r^2\right)}{\arctan\left(\frac{1}{r^2}\right)}= \frac{0}{\left(\frac{\pi}{2}\right)}=0$$
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