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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?

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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 $[-\pi/2,\pi/2]$

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

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  • $\begingroup$ The condition of $x\geq 0$ was not given. The domain of $f$ was not specified so I assume it is $\mathbb{R}^2$ $\endgroup$ – Yagger Aug 9 '18 at 18:22
  • $\begingroup$ @Yagger For $x<0$ the function is not defined therefore we need to exclude that values for the calculation of the limit. Therefore we can assume $x\ge 0$. $\endgroup$ – user Aug 9 '18 at 18:25
  • $\begingroup$ @Yagger The case is similaro to $\lim_{x\to 0} \log x$ for which we are implicitely assuming that $x\to 0^+$. For $x \to 0^-$ the expression is undefined and therefore it should be meaningless to take the limit. $\endgroup$ – user Aug 9 '18 at 18:26
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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.

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