TITLE: What does $\lim\limits_{(x,y)\rightarrow0}$ mean and how to show $ \lim\limits_{(x,y)\rightarrow0}\frac{x^3}{x^2+y^2}=0$?
QUESTION [7 upvotes]: Consider $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ where
$$f(x,y):=\begin{cases}
      \frac{x^3}{x^2+y^2} & \textit{ if }  (x,y)\neq (0,0) \\
      0 & \textit{ if }  (x,y)= (0,0)
    \end{cases} $$
If one wants to show the continuity of $f$, I mainly want to show that 
$$ \lim\limits_{(x,y)\rightarrow0}\frac{x^3}{x^2+y^2}=0$$
But what does $\lim\limits_{(x,y)\rightarrow0}$ mean? Is it equal to $\lim\limits_{(x,y)\rightarrow0}=\lim\limits_{||(x,y)||\rightarrow0}$ or does it mean $\lim\limits_{x\rightarrow0}\lim\limits_{y\rightarrow0}$?
If so, how does one show that the above function tends to zero?

REPLY [3 votes]: $$\lim_{(x,y)\rightarrow (0,0)}f(x,y)=L$$
means that for all $\epsilon>0$ there exists a  $\delta>0$ such that 
$$0<\sqrt {x^2+y^2}<\delta \implies |f(x,y)-L|< \epsilon$$
In your case let $\delta=\epsilon$.
$$\left|\frac{x^3}{x^2+y^2}\right|=|x|\cdot \left|\frac{x^2}{x^2+y^2}\right| \leq |x|\cdot1=|x|\leq \sqrt{x^2+y^2}<\delta$$
Now we can see that the aforementioned implication holds.