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?

No need to transform into polar form.
Put, $x^2+y^2=z$. Then , $z\to 0$ as $(x,y)\to (0,0)$.
Now, $$\lim_{(x,y)\to (0,0)}\frac{\tan (x^2+y^2)}{\arctan \left(\frac{1}{x^2+y^2}\right)}$$
$$=\lim_{z\to 0}\frac{\tan z}{\arctan (1/z)}=0.$$
