<s>[INST] Finding the limit of a function with 2 variables Please help to solve the limit.
$$\lim_{(x,y)\rightarrow(0,0)}\dfrac{x^3+y^3}{x^2+y^2}$$
I tried to solve it but...
$$x:=0 \Rightarrow \dfrac{x^3+y^3}{x^2+y^2} = y$$
$$y:=0 \Rightarrow \dfrac{x^3+y^3}{x^2+y^2} = x$$
 [/INST] Hints: (i) If $x^2+y^2$ is involved, switching to polar can be useful. Let $x=r\cos\theta$ and $y=r\sin\theta$.
(ii) If one does not wish to use the approach (i), note that
$$|x^3+y^3|=|x+y|(x^2-xy+y^2)\le |x+y|(2x^2+2y^2).$$
(iii) Alternately, let $\epsilon \gt 0$. If $(x,y)$ is close enough to $(0,0)$, then $|x|\lt \epsilon$ and $|y|\lt \epsilon$. Thus $|x^3|\lt \epsilon x^2$ and $|y^3|\lt \epsilon y^2$, and therefore 
$$\frac{|x^3+y^3|}{x^2+y^2}\le \frac{|x^3|+|y^3|}{x^2+y^2}\lt \frac{\epsilon x^2+\epsilon y^2}{x^2+y^2}.$$
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