Find $\lim_{(x,y) \to (0,0)} \frac{x^3+y^3}{x^2+y^2}$ assuming it exists. 
Find $\lim_{(x,y) \to (0,0)} \frac{x^3+y^3}{x^2+y^2}$ assuming it
  exists.

Since limit exists, we can approach from any curve to get the limit...
if we approach (0,0) from y=x
$\lim_{(x,y) \to (0,0)} \frac{x^3+y^3}{x^2+y^2} \Rightarrow \lim_{x \to 0} \frac{x^3+x^3}{x^2+x^2} = x = 0$ 
is this method correct?
 A: A better way to solve this problem is to use polar coordinates
$$x=r\cos \phi$$
$$y=r\sin \phi$$
$$\lim_{(x,y) \to (0,0)} \frac{x^3+y^3}{x^2+y^2}=\lim_{r\to 0}\frac{r^3(\cos^3 \phi+\sin^3 \phi)}{r^2}=\lim_{r\to 0}\left[r(\cos^3 \phi+\sin^3 \phi)\right].$$
Because $|\cos^3 \phi + \sin^3\phi|\leq 1$ the limit exists (by the sandwich theorem) and is zero. 
Remark: If you assume that the limit exists then you could also choose $y=0$ and let $x \to 0$ (because it does not matter how you approach $(0,0)$).
A: Setting $$y=tx$$ then we get
$$\frac{x^3+t^3x^3}{x^2+t^2x^2}=x\frac{1+t^3}{1+t^2}$$ this tends to Zero if $x$ tends to zero
A: You want to prove that for every $\epsilon>0$ there exists $\delta>0$ such that
$$0<\sqrt{x^2+y^2}<\delta\implies\left|\frac{x^3+y^3}{x^2+y^2}\right|<\epsilon.$$
Let $\delta=\frac{\epsilon}{2}$. Then
$$\left|\frac{x^3+y^3}{x^2+y^2}\right|\leq\left|\frac{x^3}{x^2+y^2}\right|+\left|\frac{y^3}{x^2+y^2}\right|$$
$$=|x|\underbrace{\left|\frac{x^2}{x^2+y^2}\right|}_{\leq 1}+|y|\underbrace{\left|\frac{y^2}{x^2+y^2}\right|}_{\leq 1}$$
$$\leq|x|+|y|\leq\sqrt{x^2+y^2}+\sqrt{x^2+y^2}$$
$$=2\sqrt{x^2+y^2}<2\cdot\frac{\epsilon}{2}=\epsilon$$
This proves that the limit exists and is equal to $0$.
A: Since
\begin{align}
\left|\frac{x^3+y^3}{x^2+y^2}\right|&=\left|(x+y)\cdot\frac{x^2-xy+y^2}{x^2+y^2}\right|\\
&=|x+y|\left|1-\frac{xy}{x^2+y^2}\right|\\
&\le|x+y|\left(1+\frac{|xy|}{x^2+y^2}\right)\quad(\because\text{the triangle inequality})\\
&\le|x+y|\left(1+\frac{1}{2}\right)\quad(\because\text{the AM-GM inequality})\\
&=\frac{3}{2}|x+y|
\end{align}
and $\displaystyle \lim_{(x,y)\to(0,0)}\frac{3(x+y)}{2}=0$, $\displaystyle \lim_{(x,y)\to(0,0)}\frac{x^3+y^3}{x^2+y^2}=0$ by the Squeeze Theorem.
