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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

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

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