Find the limit: $\lim\limits_{(x,y)\to (0,0)}\frac{x^3 - y^3}{x^2+y^2}$ We seek a solution to the limit question: $$\lim\limits_{(x,y)\to (0,0)}\frac{x^3 - y^3}{x^2+y^2}$$
I have an approach and think I have an answer but wanted to ask the community first. My approach is similar to the approach BabyDragon used at this link: What is $\lim_{(x,y)\to(0,0)} (x^3+y^3)/(x^2-y^2)$?
 A: We have
$$0\le\left|\frac{x^3-y^3}{x^2+y^2}\right|\le \frac{|x|x^2+|y|y^2}{x^2+y^2}\le|x|+|y|\to0$$
so the desired limit is $0$
A: What about using some polar coordinates?
$$x=r\cos t\;,\;\;y=r\sin t\implies\begin{cases} x^3-y^3=r^3(\cos t-\sin t)\left(1+\frac12\sin2t\right)\\{}\\(x,y)\to(0,0)\iff r\to 0\end{cases}$$
$$\frac{x^3-y^3}{x^2+y^2}=\frac{r^3\left(\cos^3t-\sin^3t\right)}{r^2}=r(\cos t-\sin t)\left(1+\frac12\sin2t\right)\xrightarrow[r\to 0]{}0$$
A: Seek paths to $(x,y) \to (0,0)$ by letting $y = f(x)$ where $f(0) = 0$. We can rewrite our function: $$\lim\limits_{(x,f(x))\to (0,0)}\frac{x^3 - f^3}{x^2+f^2}$$
Taking the limit using l'Hôpital's Rule we find:
$$\lim\limits_{(x,y)\to (0,0)}\frac{3x^2 - 3f^2\cdot f'}{2x+2f\cdot f'}$$
Taking the limit using l'Hôpital's Rule again:
$$\lim\limits_{(x,y)\to (0,0)}\frac{6x - \left(6f\cdot f'+ 3f^2f''\right)}{2\left(1+(f')^2 + f\cdot f''\right)}$$
We may now substitute the constraint that $f(0) = 0$ and see if the denominator is zero at any point by seeking roots to the differential equation therein. We see that it is given by $$0 = 1 + (f')^2$$ However, since $x,y \in \mathbb{R}$, the denominator cannot be zero along any path (i.e. there is no function, $f\in \mathbb{R}$, $\frac{df}{dx} = \sqrt{-1}$). Therefore, the limit exists and since $f(0) = 0$, the limit must be zero: $$\lim\limits_{(x,y)\to (0,0)}\frac{x^3 - y^3}{x^2+y^2} = 0 $$
I like this approach because of its generality. What do you think?
A: Separate the direction $[a,b]'$ from the scale $t$ and then use l'Hopital, so write $x=ta$ and $y=tb$ which gives $t \frac{a^3-b^3}{a^2+b^2} \to 0$ as $t\to 0$.
A: Expanding on Harvey Ryan Johnson's answer (give him credit also, please!)
In problems of this nature, assume $y = K x$ and see if the limit as $x\to0$ is independent of $K$. If so you have the answer. In your case,
$$
\frac{x^3-y^3}{x^2+y^2} = x \frac{1-K^3}{1+K^2} \to 0
$$
