Vtyjkyuk

I came across the following limit

$$lim_(x,y)to(0,0) fracy sqrtxsqrtx^2 + y^2$$

If $xgeq0$ then

$$0 leq bigg| fracy sqrtxsqrtx^2 + y^2bigg|= fracysqrtx^2 + y^2 leq sqrtx$$

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?

With the given condition $xge 0$, 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 $xge 0$ otherwise $f(x,y)$ is not defined and take the limit would be meaningless.

As an alternative by polar coordinates we have

$$fracy sqrtxsqrtx^2 + y^2=sinthetasqrtrcos thetato 0$$

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

$$0le |sinthetasqrtrcos theta|le sqrt r to 0$$

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.