Does a function have to be defined in a neighborhood of a point in order for a limit to exist?

Question

I came across the following limit

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

If $x\geq0$ then

$$0 \leq \bigg| \frac{y \sqrt{x}}{\sqrt{x^2 + y^2}}\bigg|= \frac{|y| \sqrt{x}}{\sqrt{x^2 + y^2}} \leq \sqrt{x}$$

So using the squeeze theorem the limit is equal to zero.

However for any neighborhood $V$ of $(0,0)$ there are points where $x \lt 0$ and therefore $f$ is not defined . Does it mean that the limit doesn't exist, or should we only consider the points where $f$ is defined?

Answer 1

With the given condition $x\ge 0$, necessary in order to have a defined expression, the limit exists and it is equal to zero.

In this case we are considering $(x,y)\to(0,0)$ along paths with $x\ge 0$ otherwise $f(x,y)$ is not defined and take the limit would be meaningless.

As an alternative by polar coordinates we have

$$\frac{y \sqrt{x}}{\sqrt{x^2 + y^2}}=\sin\theta\sqrt{r\cos \theta}\to 0$$

indeed for \theta \in $[-\pi/2,\pi/2]$

$$0\le |\sin\theta\sqrt{r\cos \theta}|\le \sqrt r \to 0$$

Answer 2

A function must be defined on an appropriate domain.

If you assume the domain must be $\mathbb R^2,$ then you do not have a function at all, let alone a limit.

If you define a domain on which $f$ is defined, and the domain contains $(0,0)$ (or at least has it as a limit point), then according to at least some definitions of the limit of a function, the limit exists.