# Limit of 2 variables function with a parameter

Find for which $$\alpha \in \Re$$ the function is continuous in $$(0,0)$$

$$f(x,y) = \begin{cases} \dfrac{-2x^3 \arctan(y)}{(x^2+y^2)^\alpha}, & (x,y) \neq (0,0) \\ 0 & (x,y) = (0,0) \end{cases}$$

To solve I need to find the value of $$\alpha$$ such that:

$$\lim_{(x,y) \to (0,0)} \frac{-2x^3 \arctan(y)}{(x^2+y^2)^\alpha} = f(0,0) = 0$$

To solve it I used the polar coordinates:

$$\left|\frac{-2\rho^3 \cos^3(\theta) \arctan(\rho \sin(\theta))}{\rho^{2\alpha}} \right|=\\= |-2\rho^{3-2\alpha} \cos^3(\theta) \arctan(\rho \sin(\theta)) | \le |-2\rho^{3-2\alpha} \arctan(\rho) |$$

Now $$\rho \to 0$$ so I can use Taylor and write:

$$|-2\rho^{3-2\alpha} \arctan(\rho) | =|-2\rho^{3-2\alpha} \rho | =|-2\rho^{4-2\alpha} |$$

So I would say that the function is continuous only if $$4 - 2\alpha \gt 0 \Leftrightarrow \alpha \lt 2$$

I tried to solve the limit with Wolfram Alpha but I am not sure if my result is right, so I would like to know if I solved it correctly

I would say you are correct; you have $$\lim_{\rho\to0} g(\rho)=\lim_{\rho\to0}\left|2p^{3-2\alpha}\arctan\rho\right|=\lim_{\rho\to0}2\rho^{4-2\alpha}=0 \quad\text{ if }\alpha\le2$$