This is the full exercise:
Given function:
$$f(x, y) = \begin{cases} \frac{|x|^\alpha y}{x^2 + y^2}, & \text{if }(x,y) \ne (0, 0) \\ 0, & \text{if } (x,y) = (0,0) \\ \end{cases}$$
a. find all $\alpha$ such that $f$ continous at $(0, 0$).
I have seen this solutions in many example, and I tried it here:
In order that $f(0, 0)$ will be continous, then $\lim_{(x, y) \to (0, 0)} \frac{|x|^\alpha y}{x^2 + y^2} = 0.$
Write $x = r\cos\theta, y = r\sin\theta$ where $0 \le \theta \le 2\pi$ and $r \to 0$.
$$\lim_{(x, y) \to (0, 0)} \frac{|r|^\alpha|\cos^\alpha\theta| r\sin\theta}{r^2(\cos^2\theta + \sin^2\theta)} = \lim_{(x, y) \to (0, 0)} \frac{r|r|^\alpha|\cos^\alpha\theta|\sin\theta}{r^2} \implies \alpha > 1.$$
The answers sheet confirms my answer,
But I wonder whether it's a valid way to find the limit or I just had luck?
Thank you very much!!