Evaluate $\lim\limits_{(x,y)\to (0,0)}\frac{\sin(xy^2)}{x}$

Question

Evaluate $$\lim_{(x,y)\to (0,0)}\frac{\sin(xy^2)}{x}$$

$$\lim_{(x,y)\to (0,0)}\frac{\sin(xy^2)}{x}=\lim_{(x,y)\to (0,0)}\frac{\sin(xy^2)}{xy^2}y^2$$

now $$\lim_{(x,y)\to (0,0)}\frac{\sin(xy^2)}{xy^2}\cdot \lim_{(x,y)\to (0,0)} y^2=1\cdot 0=0$$

1. Is there a problem with this calculation regarding the where the function is defined?

2. In general if we look at $(x,0)\to(0,0)$ or $(0,y)\to(0,0)$ are those iterative limits?

Answer 1

Note that, since for $x=0$ the expression is not defined, the limit does not exist.

If we define $f(x,y)=0$ for $x=0$ then by squeeze theorem

$$0<\frac{\sin(xy^2)}{x}=\frac{\sin(xy^2)}{xy^2}y^2<y^2\to 0$$

the limit exists and is equal to $0$.