Choosing two distinct paths for calculating the limits of two-variable functions

Question

Currently I have been learning the limits of two-variable functions. I know that in order to show the non-existence of a given limit, we need to select two distinct paths for testing. If the two outcomes are different, the limit does not exist. Yet, I don't know the exact way for path selection. To be more specific, let's refer to the example below:

$\lim_{(x,y) \to (0,0)} \frac{x^3y}{x^6+y^2}$

Originally I chose $\ y=mx$ and $\ y=mx^2$. When $\ y=mx$,

$\lim_{(x,y) \to (0,0)} \frac{x^3y}{x^6+y^2}$

=$\lim_{(x,mx) \to (0,0)} \frac{x^3(mx)}{x^6+(mx)^2}$

=$\lim_{(x,mx) \to (0,0)} \frac{mx^4}{x^6+m^2x^2}$

=$\lim_{(x,mx) \to (0,0)} \frac{mx^4}{x^2(x^4+m^2)}$

=$\lim_{(x,mx) \to (0,0)} \frac{mx^2}{x^4+m^2}$

=$\frac{0}{0+m^2}$

=$\ 0$

When $\ y=mx^2$,

$\lim_{(x,y) \to (0,0)} \frac{x^3y}{x^6+y^2}$

=$\lim_{(x,mx^2) \to (0,0)} \frac{x^3(mx^2)}{x^6+(mx^2)^2}$

=$\lim_{(x,mx^2) \to (0,0)} \frac{mx^5}{x^6+m^2x^4}$

=$\lim_{(x,mx^2) \to (0,0)} \frac{mx^5}{x^4(x^2+m^2)}$

=$\lim_{(x,mx^2) \to (0,0)} \frac{mx}{x^2+m^2}$

=$\frac{0}{0+m^2}$

=$\ 0$

Therefore for these two paths the outcomes appear to be the same. Yet, I also tried $\ y=mx^3$, and the outcome does not equal to zero. So eventually I successfully proved the non-existence of the given limit. But is there any efficient way for path selection so that I don't need to randomly choose two paths relating variables $\ x$ and $\ y$?

Answer 1

$$\lim_{(x,y) \to (0,0)} \;\frac{x^3y}{x^6+y^2}$$

There is no one single heuristic that works in all cases. But for this case in particular, once you had the insight already to check the paths $y = x$ and $y=x^2\,$, you may have recognized that the limit was $0\,$ along those paths because the denominator still had terms in $x$ of lower degree than the numerator. Next path to investigate would be whether there is an exponent $\alpha$ for which both $x^3y^\alpha$ and $x^6+y^{2\alpha}$ equal the same power of $x$ i.e. $3+\alpha=6=2 \alpha$ , so that the fraction reduces to a ratio independent of $x$. It (ever so luckily) happens that $\alpha=3$ fits the bill, and the limit along the curve $y=x^3$ is $\frac{1}{2}\,$, which proves that the general limit does not in fact exist.