Does the limit $\lim\limits_{(x,y)\to (0,0)}\frac{x^3-y^3}{3x+2y}$ exist?

Question

Does this limit exist? $$\lim\limits_{(x,y)\to (0,0)}\frac{x^3-y^3}{3x+2y}$$

From what I understand, the line $3x+2y=0$ goes through $(0,0)$, so the limit doesn't exist. But shouldn't I be able to find two paths with different limits in that case? I can't seem to find them.

Answer 1

Note that the points such that $3x+2y=0$ are simply excluded from the domain and thus we can't conclude that the limit doesn't exist from here.

To show that we need to find two different paths with different limits as $(x,y)\to (0,0)$.

Notably in this case we have that

• for $x=y$ we have $\dfrac{x^3-y^3}{3x+2y}=0$

• for $x=t\to 0$ and $y=-\frac32t+t^3$ we have $=\dfrac{x^3-y^3}{3x+2y}=\dfrac{t^3-(-\frac32t+t^3)^3}{3t-3t+2t^3}\to L\in \mathbb{R}\setminus\{0\}$

Answer 2

Go along curves $y=ax+bx^3$ for suitable $a$ and $b$.