Multivariable Calculus, Two Path Test.

Question

I'm have trouble understanding how you determine which paths you should choose.

In the book for the function lim as (x,y) goes to (0,0) $$f(x,y) = \frac{2x^2y}{x^4+y^2}$$ They say "We examine the values of $f$ along curves that end at $(0,0)$. Along the curve $y=kx^2$, x cannot equal 0."

I don't understand the above quote and how did they come up with $y = kx^2$.

It makes our ratio look nice (cancellation). – André Nicolas Dec 07 '13 at 18:39 — André Nicolas, Dec 07 '13 at 18:39
Besides, you know the curve $y=kx^2$ ends in $(0,0)$ – user99680 Dec 07 '13 at 20:04 — user99680, Dec 07 '13 at 20:04
In general you should try $y=kx^\alpha, \alpha>0$. – Vim Jun 22 '16 at 07:10 — Vim, Jun 22 '16 at 07:10

score 3 · Answer 1 · answered Dec 07 '13 at 18:39

There is no systematic approach to considering curves along which the limit might assume distinct values. This example is a very particular one: among all lines passing through the origin it is zero, but among any parabola it takes another value.

This is one of those things where experience is handy: solving many problems and tackling with different approaches is what guides us.

Perhaps the intuition for taking a parabola in this case is noticing that the exponent of $y$ is exactly two times the $x$. You can construct many similar ones in this way:

$$f(x,y) = \frac{7x^3 y}{x^6 + y^2}, \quad f(x,y) = \frac{13x^5y}{x^{10}+y^2},$$

just to create a few. In these cases you'll find the same: along lines it is zero, along curves such as $y=x^3$ and $y=x^5$ each will have a different limit.

Multivariable Calculus, Two Path Test.

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